chem11_sm_9_5

Section 9.5: Stoichiometry of Solutions Tutorial 1 Practice, page 447 1. (a) Given: VSrCl = 150 mL 2 cSrCl = 0.25 mol / L 2 cNa CO = 0.500 mol/L 2 3 Required: volume of 0.500 mol/L sodium carbonate, VNa CO 2 3 Solution: Step 1. Convert all volumes of the solutions to litres. 1L VSrCl = 150 mL ! 2 1000 mL VSrCl = 0.150 L 2 Step 2. Write a dissociation equation listing the calculated amounts and the required value(s). SrCl2(aq) + Na2CO3(aq) → SrCO3(s) + NaCl(aq) VSrCl = 0.150 L cNa CO = 0.500 mol/L 2 2 cSrCl = 0.25 mol / L 2 3 VNa CO 2 3 Step 3. Determine the amount of strontium chloride by rearranging the equation. n c= V nSrCl = cSrCl VSrCl 2 2 2 0.25 mol ! 0.15 L 1L = 0.037 50 mol [2 extra digits carried] = nSrCl 2 Step 4. Determine the amount of sodium carbonate. 1 mol Na CO 2 3 nNa CO = 0.037 50 molSrCl ! 2 3 2 1 molSrCl 2 nNa CO = 0.037 50 mol 2 3 Step 5. Determine the volume of sodium carbonate by rearranging the equation. nNa CO 2 3 VNa CO = 2 3 cNa CO 2 = 3 0.037 50 mol 0.500 mol L VNa CO = 0.075 L 2 3 Statement: The volume of 0.500 mol/L sodium carbonate required is 0.075 L or 75 mL. Copyright © 2011 Nelson Education Ltd. Chapter 9: Solutions and Their Reactions 9.5-1 037 50 mol (from (a)) 2 Required: mass of precipitate strontium carbonate.037 50 mol 2 Step 2.5 g 3 Statement: The mass of strontium carbonate precipitate is 5.0 mL ! 1000 mL VKOH = 0. Determine the amount of precipitate of strontium carbonate.0 L Required: mass of iron(III) hydroxide precipitate.50 mol/L = 40. 147. 1 molSrCO 3 nSrCO = 0. mFe(OH) 3 Solution: Step 1. Calculate the mass of precipitate of strontium carbonate. Fe(NO3)3(aq) + 3 KOH (aq) → Fe(OH)3(s) + 3 KCl(aq) cKOH = 0.80 mol / L cFe(NO ) = 0.070 L Copyright © 2011 Nelson Education Ltd.040 L 3 3 Step 2.50 mol/L mFe(OH) 3 3 VFe(NO 3 )3 3 = 0.80 mol / L cFe(NO 3 )3 VFe(NO 3 )3 = 0.0 mL cKOH = 0.5-2 . Write the dissociation equation listing the calculated amounts and the required value(s). Chapter 9: Solutions and Their Reactions 9.0 mL ! 3 3 1L 1000 mL VFe(NO ) = 0. 1L VKOH = 70. (a) Given: VKOH = 70. Convert all volumes of the solutions to litres.(b) Given: nSrCl = 0.037 50 mol ! 3 1 mol mSrCO = 5.040 L VKOH = 0.070 L VFe(NO ) = 40.037 50 molSrCO 3 3 Step 3. Write a dissociation equation listing the calculated amounts and the required value(s).037 50 molSrCl ! 3 2 1 molSrCl 2 nSrCO = 0. SrCl2(aq) + Na2CO3(aq) → SrCO3(s) + NaCl(aq) nSrCl = 0.63 g mSrCO = 0.5 g. 2. mSrCO 3 Solution: Step 1. 056 mol KOH ! 3 3 3 mol KOH nFe(NO 3 )3 = 0.040 L 1L = 0.056 mol potassium hydroxide requires 0.056 mol = nKOH Step 4. 1 mol Fe(NO ) 3 3 nFe(NO ) = 0. 0. Step 5. Copyright © 2011 Nelson Education Ltd.0 g 3 Statement: The mass of iron(III) hydroxide precipitate is 2. Since 0.020 mol = nFe(NO 3 )3 nKOH = cKOHVKOH 0. Determine the limiting reagent.Step 3.0187 mol Fe(OH) [1 extra digit carried] 3 3 Step 6. Use the amount of limiting reagent to determine the amount of iron(III) hydroxide precipitate. Chapter 9: Solutions and Their Reactions 9.056 mol KOH ! 3 3 mol KOH nFe(OH) = 0. potassium hydroxide is the limiting reagent. 1 mol Fe(OH) 3 nFe(OH) = 0.0187 mol ! 3 1 mol mFe(OH) = 2.88 g mFe(OH) = 0. Determine the amounts.070 L 1L = 0.0187 mol Fe(NO 3 )3 [1 extra digit carried] Therefore.5-3 .80 mol ! 0. n c= V nFe(NO ) = cFe(NO ) VFe(NO ) 3 3 3 3 3 3 0.50 mol ! 0.0 g. Determine the mass of iron(III) hydroxide precipitate. 106.019 mol of iron(III) nitrate.020 mol of iron(III) nitrate is present. cOH – = 6. Chapter 9: Solutions and Their Reactions 9. (a) Given: cBaCl = 0. Convert the concentration of the compound into concentration of anions.30 mol/L.50 ! L 1 mol BaCl 2 cCl– = 1. cOH – Solution: Step 1: Write the dissociation reaction. cClO– 3 Solution: Step 1: Write the dissociation reaction. mol BaCl 2 molCl– 2 cCl– = 0. Convert the concentration of the compound into concentration of anions. KOH(s) → K+(aq) + OH–(aq) Step 2. Copyright © 2011 Nelson Education Ltd.30 mol/L 3 Statement: The amount concentration of chlorate anion is 0.0 mol/L.0 mol KOH L ! 1 molOH – 1 mol KOH cOH – = 6. Al(ClO3)3(s) → Al3+(aq) + 3 ClO3–(aq) Step 2. Convert the concentration of the compound into concentration of cations. (b) Given: cKOH = 6.0 mol/L Statement: The amount concentration of chloride anion is 1.10 ! 3 L 1 molAl(ClO ) 3 3 cClO– = 0.10 mol/L 3 3 Required: amount concentration of the anion.5-4 .0 mol/L. (c) Given: cAl(ClO ) = 0.50 mol/L 2 Required: amount concentration of the anion. BaCl2(s) → Ba2+(aq) + 2 Cl–(aq) Step 2. cCl– Solution: Step 1: Write the dissociation reaction.0 mol/L Statement: The amount concentration of hydroxide anion is 6. page 448 1.Tutorial 2 Practice.0 mol/L Required: amount concentration of the anion. molAl(ClO ) 3 molClO– 3 3 3 cClO– = 0. Determine the amount concentration.11 g n(NH 4 ) 2 CO3 = 0. Chapter 9: Solutions and Their Reactions 9.1000 L = 1.50 mol/L 4 2 3 Step 5.0 mL 4 ) 2 CO3 Required: amount concentration of ammonium ion. cNH + 4 Solution: Step 1.0 mL ! 4 2 3 1000 mL V(NH ) CO = 0.149 83 mol [2 extra digits carried] Step 3.50 ! 4 L 1 mol(NH ) CO 4 2 3 cNH + = 3.4 g ! 4 2 3 96. (NH4)2CO3(s) → 2 NH4+(aq) + CO32–(aq) c(NH ) CO = 1.5-5 .4 g = 100. Determine the amount of solute.2. 1 mol n(NH ) CO = 14. n(NH ) CO 4 2 3 c(NH ) CO = 4 2 3 V(NH ) CO 4 2 3 0.1000 L 4 2 3 Step 2.00 mol/L 4 Statement: The amount concentration of the ammonium ion is 3. Copyright © 2011 Nelson Education Ltd. Given: m(NH V(NH 4 ) 2 CO3 = 14. mol(NH ) CO 2 mol NH + 4 2 3 4 cNH + = 1.00 mol/L.50 mol/L = c(NH 4 ) 2 CO3 Step 4. Convert the concentration of the compound into concentration of cations. Write a dissociation equation listing the calculated amounts and the required value(s).149 83 mol 0. Convert the volume of the solution to litres. 1L V(NH ) CO = 100. 3. Determine the amount of solute. 1L VAgNO = 32.75 L 3 4 mNa PO 3 4 Step 2. n c= V nNa PO = cNa PO VNa PO 3 4 3 4 3 4 0.083 mol/L 3 4 Step 3.083 mol ! 1. Chapter 9: Solutions and Their Reactions 9.5 Questions. Given: VNa PO = 1.75 L 1L = 0. 163. Na3PO4(s) → 3 Na+(aq) + PO43–(aq) cNa + = 0. page 449 1. Convert the concentration of the sodium ion into concentration of sodium phosphate.100 mol / L 3 VNaCl = 25 mL Required: amount concentration of sodium ions. Determine the mass of sodium phosphate.032 L 3 Copyright © 2011 Nelson Education Ltd.5-6 .0 mL 3 cAgNO = 0. Write a dissociation equation listing the calculated amounts and the required value(s).25 ! 3 4 L 3 mol Na + cNa PO = 0. mNa PO 3 4 Solution: Step 1. mol Na + 1 mol Na PO 3 4 cNa PO = 0.75 L 3 4 cNa + = 0.146 mol [1 extra digit carried] = nNa PO 3 4 Step 4.0 mL ! 3 1000 mL VAgNO = 0.146 mol ! 3 4 1 mol mNa PO = 24 g 3 4 Statement: The mass of sodium phosphate is 24 g. Convert all volumes of the solutions to litres. (a) Given: VAgNO = 32.25 mol / L VNa PO = 1.25 mol / L Required: mass of sodium phosphate. cNa + Solution: Step 1.95 g mNa PO = 0. Section 9. VNaCl = 25 mL ! 1L 1000 mL VNaCl = 0.025 L Step 2. Chapter 9: Solutions and Their Reactions 9.5-7 . Determine the amount concentration of sodium chloride. NaCl(aq) → Na+(aq) + Cl–(aq) Step 7.003 20 mol [1 extra digit carried] = nAgNO 3 Step 4.003 20 mol NaCl Step 5.13 mol NaCl L ! 1 mol Na + 1 mol NaCl cNa + = 0.003 20 mol 0. Determine the amount of solute. 1 mol NaCl nNaCl = 0. n cNaCl = NaCl VNaCl 0. cNa + = 0. Write the dissociation reaction for sodium ions. Copyright © 2011 Nelson Education Ltd.032 L VNaCl = 0.100 mol ! 0. n c= V nAgNO = cAgNO VAgNO 3 3 3 0.003 20 molAgNO ! 3 1 molAgNO 3 nNaCl = 0. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) VAgNO = 0.025 L = 0.13 mol/L Statement: The amount concentration of sodium ions is 0. Convert the concentration of the compound into concentration of sodium cations.13 mol/L.100 mol / L 3 cNaCl Step 3.032 L 1L = 0.13 mol/L = cNaCl Step 6.025 L 3 cAgNO = 0. Write a dissociation equation listing the calculated amounts and the required value(s). Determine the amount of sodium chloride. 0250 L 4 cNiSO = 0.0 mL cNaOH = 1.54 g mNaCl = 0.0 mL 4 VNaOH = 25. Convert all volumes of the solutions to litres.032 L nNaCl = 0.0 mL ! 1L 1000 mL VNaOH = 0. Chapter 9: Solutions and Their Reactions 9. 2.5 g/L.5 g/L = cNaCl Statement: The amount concentration of sodium chloride is 7. mNi(OH) 2 Solution: Step 1.187 g [1 extra digit carried] Step 2.0500 L VNaOH = 0. Convert the amount of solute to mass.0250 L Step 2.003 20 mol ! 1 mol mNaCl = 0.0 mL ! 4 1000 mL VNiSO = 0.(b) Given: VNaCl = 0.00 mol/L Copyright © 2011 Nelson Education Ltd. 1 extra digit carried) Required: amount concentration of sodium chloride in g/L Solution: Step 1. 1L VNiSO = 50.5-8 .45 mol / L 4 cNaCl = 1.187 g 0. (a) NiSO4(aq) + 2 NaOH(aq) → Ni(OH)2(s) + Na2SO4(aq) (b) Given: cNiSO = 0.0500 L 4 VNaOH = 25.45 mol / L 4 VNiSO = 50. 58.00320 mol / L (from (a).025 L = 7.00 mol / L Required: mass of nickel hydroxide precipitate. Determine concentration of the solute using the mass. m cNaCl = NaCl VNaCl 0. NiSO4(aq) + 2 NaOH(aq) → Ni(OH)2(s) + Na2SO4(aq) VNiSO = 0. Write a dissociation equation listing the calculated amounts and the required value(s). 1 mol Ni(OH ) 2 nNi(OH ) = 0.5-9 . Convert the volume of the solution to litres.0125 mol ! 2 1 mol mNi(OH ) = 1. Step 5.0 mL 2 3 Required: amount concentration of sodium carbonate solution. n c= V nNiSO = cNiSO VNiSO 4 4 4 0.0250 mol Step 4.0 mL ! 2 3 1000 mL VNa CO = 0. sodium hydroxide is the limiting reagent and nickel(II) sulfate is the excess reagent.0250 L 1L nNaOH = 0. Determine the amount of sodium hydroxide required to react with 0. Determine the mass of nickel hydroxide precipitate. Determine the amount of nickel hydroxide precipitate.16 g 2 Statement: The mass of nickel(II) hydroxide precipitate produced in this reaction is 1.71 g mNi(OH ) = 0. cNa CO 2 3 Solution: Step 1. 3.0225 mol [1 extra digit carried] = nNiSO 4 nNaOH = cNaOHVNaOH 1. 2 mol NaOH nNaOH = 0.0250 mol NaOH ! 2 2 mol NaOH nNi(OH ) = 0.Step 3. Determine the amounts of both substances.0450 mol Since this amount is greater than the amount present.2000 L 2 3 Copyright © 2011 Nelson Education Ltd. 92.0500 L 1L = 0. Chapter 9: Solutions and Their Reactions 9.16 g.0225 mol NiSO ! 4 1 mol NiSO = 4 nNaOH = 0.00 mol ! 0. 1L VNa CO = 200.2 g 3 VNa CO = 200.45 mol ! 0.0225 mol of nickel(II) sulfate solution. (a) Given: mCaCO = 15.0125 mol Ni(OH ) 2 2 Step 6. 1 molCaCl 2 nCaCl = 0.759 mol/L. Determine the amount of calcium chloride. Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2 NaCl(aq) VNa CO = 0.759 mol/L = cNa CO 2 3 Statement: The amount concentration of original sodium carbonate solution is 0. Determine the amount of calcium carbonate.151 86 mol 0.200 L mCaCO = 15.151 86 mol (from (a).Step 2. Determine the amount concentration. VCaCl Solution: Step 1.2000 L = 0.151 86 mol 2 3 Step 5.500 mol/L 2 nCaCO = 0. 1 mol Na CO 2 3 nNa CO = 0.2 g 2 3 3 nCaCO = 0. Chapter 9: Solutions and Their Reactions 9.500 mol/L mCaCO = 15.151 86 mol [2 extra digits carried] 3 Step 3. nNa CO 2 3 cNa CO = 2 3 VNa CO 2 3 0.2 g ! 3 100. 1 mol nCaCO = 15.2 g 3 cCaCl = 0. Determine the amount of sodium carbonate. Write a dissociation equation listing the calculated amounts and the required value(s). 2 extra digits carried) 3 Required: volume of calcium chloride solution.151 86 mol 3 Step 2.09 g nCaCO = 0. Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2 NaCl(aq) cCaCl = 0.2 g 2 3 nCaCO = 0.151 86 molCaCO ! 2 3 1 molCaCO 3 nCaCl = 0.151 86 mol 2 Copyright © 2011 Nelson Education Ltd.151 86 mol 3 Step 4.151 86 molCaCO ! 2 3 3 1 molCaCO 3 nNa CO = 0. (b) Given: mCaCO = 15. Write a dissociation equation listing the calculated amounts and the required value(s).5-10 . 0 mol/L mFe O = 224 g 3 3 Required: volume of hydrochloric acid. Given: cHCl = 12.4026 mol [2 extra digits carried] 2 3 Step 2. Determine the amount of hydrochloric acid.2 g of calcium carbonate is 0. nCaCl 2 cCaCl = 2 VCaCl 2 VCaCl = 2 = nCaCl cCaCl 2 2 0.4026 mol 2 3 Step 3.5-11 . 1 mol nFe O = 224 g ! 2 3 159.4026 mol Fe O ! 2 3 1 mol Fe O 2 3 nHCl = 8. Rearrange the equation and substitute values. Write a dissociation equation listing the calculated amounts and the required value(s).4158 mol HCl [2 extra digits carried] Copyright © 2011 Nelson Education Ltd.7 g nFe O = 1.500 mol L VCaCl = 0. 6 mol HCl nHCl = 1.Step 3. 4. VHCl Solution: Step 1. Determine the amount of solute from the mass of solute.304 L 2 Statement: The volume of calcium chloride solution required to produce 15.0 mol/L mFe O = 224 g 2 3 nFe O = 1. Fe2O3(s) + 6 HCl(aq) → 2 FeCl3(aq) + 3 H2O(l) cHCl = 12.151 86 mol 0. Chapter 9: Solutions and Their Reactions 9.304 L or 304 mL. 15 L 4 Step 2.4158 mol mol L VHCl = 0.100 mol 1L = 0.701 L. mAl Solution: Step 1. 2 molAl nAl = 0.010 molAl Copyright © 2011 Nelson Education Ltd.701 L Statement: The volume of hydrochloric acid is 0.015 molCuSO ! 4 3 molCuSO 4 nAl = 0.015 mol = 0. Rearrange the equation and substitute values to calculate the volume of hydrochloric acid. (a) 2 Al(s) + 3 CuSO4(aq) → 3 Cu(s) + Al2(SO4)3(aq) (b) Given: cCuSO = 0. Convert the volume of the solution to litres.100 mol/L 12. 5. n cHCl = HCl VHCl VHCl = = nHCl cHCl 8.15 L ! nCuSO 4 Step 3.Step 4. Chapter 9: Solutions and Their Reactions 9.5-12 . Determine the amount of copper(II) sulfate. 1L VCuSO = 150 mL ! 4 1000 mL VCuSO = 0.0 4 VCuSO = 150 mL 4 Required: mass of aluminum. nCuSO 4 cCuSO = 4 VCuSO 4 nCuSO = VCuSO ! cCuSO 4 4 4 0. Determine the amount of aluminum. 010 mol ! 1 mol mAl = 0. Convert the mass of sodium hydroxide to grams. 26. 1 mol NaCl nNaCl = 1. 106 g 6 mNaOH = 45 ! 10 t ! 1t mNaOH = 45 ! 1012 g Step 2.125 ! 1012 mol NaCl Step 5.27 g Statement: The mass of aluminum required to react with the copper(II) sulfate solution is 0. Given: mNaOH = 45 ! 10 t cNaCl = 6. 1 mol nNaOH = 45 ! 1012 g ! 40.125 ! 1012 mol [2 extra digits carried] Step 3. Write a dissociation equation listing the calculated amounts and the required value(s). Determine the amount of sodium chloride.0 VNaCl Statement: The volume of sodium chloride solution required is 1. n cNaCl = NaCl VNaCl VNaCl = = nNaCl cNaCl 1.00 g nNaOH = 1. Copyright © 2011 Nelson Education Ltd.125 ! 1012 mol NaOH ! 1 mol NaOH nNaCl = 1. 2 NaCl(aq) + 2 H2O(l) → 2 NaOH(aq) + Cl2(g) + H2(g) nNaOH = 1. 6 6.27 g.98 g mAl = 0.0 mol/L Required: volume of sodium chloride.0 mol/L Step 4. Determine the mass of aluminum. VNaCl Solution: Step 1.9 ! 10 L 6.5-13 . Chapter 9: Solutions and Their Reactions 9.125 ! 1012 mol cNaCl = 6.125 ! 1012 mol mol L 11 = 1.9 × 1011 L. Rearrange the equation and substitute values to calculate the volume of hydrochloric acid. Determine the amount of sodium hydroxide.Step 4. 0 mL ! 3 1000 mL VAgNO = 0. determine the amount of sodium chloride.42 g Vsolution = 100.50 mol / L 3 VAgNO = 15.2 ! 10 mol nAgNO 3 Step 5. Using calcium chloride as the solute.0150 L 3 Vsolution = 100. Using sodium chloride as the solute.0 mL Required: identify the solute Solution: Step 1. 1 mol nNaCl = 0.42 g ! 2 110. Using calcium chloride as the solute.44 g nNaCl = 7. Determine the amount of silver nitrate used. Using sodium chloride as the solute.78 ! 10 "3 mol [1 extra digit carried] 2 Step 7.2 ! 10 "3 mol NaCl ! 3 1 mol NaCl nAgNO = 7. Convert the volumes of the solutions to litres.98 g nCaCl = 3. write the dissociation equation.78 ! 10 mol 2 3 Copyright © 2011 Nelson Education Ltd. 0.2 ! 10 "3 mol Step 4. CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)2(aq) "3 nAgNO nCaCl = 3. Using sodium chloride as the solute. 1 molAgNO 3 nAgNO = 7. determine the amount of calcium chloride.2 ! 10 "3 mol 3 Step 6. determine the amount of silver nitrate required. NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3 (aq) "3 nNaCl = 7.0 mL ! 1L 1000 mL Vsolution = 0.1000 L Step 2.0150 L ! 3 1L nAgNO = 7.50 mol nAgNO = 0.5-14 . Given: cAgNO = 0. write the dissociation equation.0 mL 3 msolute = 0.5 ! 10 "3 mol 3 Step 3.7.42 g ! 58. 1L VAgNO = 15. Chapter 9: Solutions and Their Reactions 9. 1 mol nCaCl = 0. (b) Given: c( NH ) SO = 0.4 mol/L.0 mol/L.0 mol/L Statement: The amount concentration of cations is 3. mol Na CO 2 mol Na + 2 3 cNa + = 0.5 mol / L 2 4 3 Step 1: Write the dissociation reaction. Na2CO3(aq) → 2 Na+(aq) + CO32-(aq) Step 2.56 ! 10 "3 mol 3 Statement: The actual amount of silver nitrate available is 7. (c) Given: cFe (SO ) = 1.5 mol / L 2 3 Step 1: Write the dissociation reaction.5-15 . it would take 7. mol(NH ) SO 2 mol NH + 4 2 3 4 cNH + = 0. Copyright © 2011 Nelson Education Ltd. determine the amount of silver nitrate required.5 ! L 1 mol Fe (SO ) 2 4 3 cFe3+ = 3. Convert the concentration of the compound into concentration of cations.5 ! L 1 mol Na CO 2 3 cNa + = 1 mol/L Statement: The amount concentration of cations is 1 mol/L.56 × 10-3 mol of silver nitrate to precipitate all of the chloride ions. Chapter 9: Solutions and Their Reactions 9. Fe2(SO4)3(s) → 2 Fe3+ + 3 SO42– Step 2. (a) Given: cNa CO = 0. If the unknown solution contains sodium chloride.2 ! 4 L 1 mol(NH ) SO 4 2 3 cNH + = 0. If the unknown solution contains calcium chloride. it would take 7.2 mol / L 4 2 3 Step 1: Write the dissociation reaction.5 ×10-3 mol. Convert the concentration of the compound into concentration of cations.4 mol/L 4 Statement: The amount concentration of cations is 0. mol Fe (SO ) 2 mol Fe3+ 2 4 3 cFe3+ = 1. The unknown solution is sodium chloride because there is not enough silver nitrate to precipitate all of the chloride ions if the solution was calcium chloride.78 ! 10 "3 molCaCl ! 3 2 1 molCaCl 2 nAgNO = 7. (NH4)2SO3(s) → 2 NH4+ + SO32– Step 2. Convert the concentration of the compound into concentration of cations. 8.Step 8. Using calcium chloride as the solute. 2 molAgNO 3 nAgNO = 3.2 × 10-3 mol of silver nitrate to precipitate all of the chloride ions. mNa CO 2 3 Solution: Step 1.85 mol/L VNa CO = 200.2000 L L = 0.99 g mNa CO = 0.425 mol/L [1 extra digit carried] 2 3 Step 4. nNa CO 2 3 cNa CO = 2 3 VNa CO 2 3 nNa CO = cNa CO ! VNa CO 2 nNa CO 2 3 3 2 3 2 3 mol = 0. Na2CO3(aq) → 2 Na+(aq) + CO32-(aq) Step 3.425 ! 0. Copyright © 2011 Nelson Education Ltd.085 mol Step 5. Determine the amount of sodium carbonate.0 g.9. Write the dissociation reaction.0 g 2 3 Statement: The mass of sodium carbonate is 9.0 mL 2 3 Required: mass of sodium carbonate. Chapter 9: Solutions and Their Reactions 9.0 mL ! 2 3 1000 mL VNa CO = 0.2000 L 2 3 Step 2. Convert the volume of the solution to litres. mol Na + 1 mol Na CO 2 3 cNa CO = 0.5-16 . 105.85 ! 2 3 L 2 mol Na + cNa CO = 0.085 mol ! 2 3 1 mo l mNa CO = 9. Given: cNa + = 0. Determine the mass of sodium carbonate. Determine the concentration of sodium carbonate. 1L VNa CO = 200.