CHEMISTRY 181 UKZN TUTORIAL 6 STOICHIOMETRY AND SOLUTION CHEMISTRY. 1. Consider the degradation of glucose to carbon dioxide and water:

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1 CHEMISTRY 8 UKZN - 20 TUTORIAL 6 STOICHIOMETRY AND SOLUTION CHEMISTRY. Consider the degradation of glucose to carbon dioxide and water: C6H2O6 + 6O2 6 CO2 + 6 H2O If 856 g of C6H2O6 is consumed by a person over a certain period of time, what is the mass of CO2 produced? Ans: Stoichiometry nc 6H 2O 6 = nco2 6 n C6H2O6 = 856 g 80.2 g/mol = mol C6H2O6 nco 2 = mol x 6 = mol CO2 Mass of CO2 = mol CO2 x 44.0 g mol =.254 x 0 3 g 2. The reaction between Li(s) and H2O(l) to produce H2(g) is: 2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g) How many grams of Li are needed to produce 9.89 g of H2(g)? Ans. Conversion steps grams of H2 moles of H2 moles of Li grams of Li nli 2 = nh2 therfore nli = nh2 x 2 = 9.89 g 2.06 g/mol x 2 = 9.82 mols Mass of Li = 9.82 mols x 6.94 g mol = 68. g

2 3. Titanium metal is prepared by the reaction: TiCl4(g) + 2 Mg(l) Ti(s) + 2MgCl2(l) In a certain industrial operation that makes aircrafts parts, g of TiCl4 are reacted with g of Mg. (a) Show using appropriate calculations which is the limiting reagent. Ans: Calculating the no of mols of product Ti from both reactants Note: the product MgCl2 can also be used instead of Ti nti = nticl4/ = 3.54 x 07 g 89.7 g mol =.87 x 05 mol nti = nmg/2 =.3 x 07 g 24.3 g mol = mol therefore TiCl4 is the limiting reagent (b) Calculate the theoretical yield of Ti in grams. Ans: Use the limiting reagent to calculate the theroretical yield nti = nticl4/ = g 89.7 g mol = mol mass of Ti = mol g mol = g (c) Calculate the percentage yield if 7.9 x 0 obtained. Ans. % Yield = actual yield 00 = 7.9 x 06 g x 00 = 88.4 % theoretical yield 8.95 x 0 6 g 6 g of Ti are actually

3 4. Calculate the mass of solute (in grams) contained in the following solutions: (a) 40.0 ml of M KCl Ans. No. of mols KCl (n) = molarity volume = M dm 3 (40.0 ml = dm 3 ) = mols Mass of KCl = ( mols ) g mol = g (b) 2.50 ml of M KMnO4 Ans. No. of mols KMnO4 (n) = molarity volume = M dm 3 = mols Mass of KMnO4 = ( mols ) g mol = g 5. What mass of salt is required to prepare 500 ml of a 00 ppm Cr solution using K2CrO4. Ans. 00 ppm = 00 mg K2CrO4 per 000 ml solvent % K in K2CrO4 = molar mass of K x 2 molar mass of K 2CrO 4 00 = 2 x 39.0 g mol 94.2 g mol x 00 = % If 00 mg of K2CrO4 is dissolved in 000 ml solvent, it only constitutes % by K. 00 mg = % K x mg = 00 % K therefore x = mg needs to be dissolved in 000 ml solvent.

4 Hence mg or.2435 g of K2CrO4 should be dissolved 500 ml of solvent to obtain a 00ppm solution. 6. A solution is prepared by dissolving 0.64g of (NH4)2Ce(NO3)6 (FW = 548.3g) to L with distilled water. Calculate the ppm of : (a) NH4 + (b) Ce 4+ (c) NO3 - (a) NH4 + - % NH4 + in (NH4)2Ce(NO3)6 = g mol g mol x 00 = 3.29 % Mass of NH4 + = g = g = 3.83 mg Therefore conc = 383 mg/l = 383 ppm (b) Ce 4+ % Ce 4+ in (NH4)2Ce(NO3)6 = 40. g mol g mol x 00 = % Mass of Ce4+ = g = x 0-3 g = mg Therefore conc = mg/l = ppm (c) NO3 - % NO3 - in (NH4)2Ce(NO3)6 = 6 x 62.0 g mol g mol x 00 = % Mass of NO3 = g = x 0-3 g = mg Therefore conc = mg/l = ppm

5 7. What volume of 6M HNO3 is required to prepare 250 ml of: (a) 0.0 M HNO3 MHNO 3(init) vol HNO 3(init) = MHNO 3(final) vol HNO 3(final) 6 M (init) vol HNO 3(init) = 0.0 M (final) 250 ml (final) vol HNO 3(init) = 0.6 ml (b) 0.5 M HNO3 MHNO 3(init) vol HNO 3(init) = MHNO 3(final) vol HNO 3(final) 6 M (init) vol HNO 3(init) = 0.5 M (final) 250 ml (final) vol HNO 3(init) = 7.8 ml (c) 6M HNO3 MHNO 3(init) vol HNO 3(init) = MHNO 3(final) vol HNO 3(final) 6 M (init) vol HNO 3(init) = 6 M (final) 250 ml (final) vol HNO 3(init) = ml 8. What is the final concentration when 50 ml of a solution of 4 M NH3 is diluted to: (a) 00 ml MNH 3(init) vol NH 3(init) = MNH 3(final) vol NH 3(final) 4 M (init) 50 ml(init) = MNH 3 (final) 00 ml (final) MNH 3 (final) = 7.0 M

6 (b) 250 ml MNH 3(init) vol NH 3(init) = MNH 3(final) vol NH 3(final) 4 M (init) 50 ml(init) = MNH 3 (final) 250 ml (final) MNH 3 (final) = 2.8 M (c) L MNH 3(init) vol NH 3(init) = MNH 3(final) vol NH 3(final) 4 M (init) 50 ml(init) = MNH 3 (final) 000 ml (final) MNH 3 (final) = 0.7 M 9. Muriatic acid (HCl) is sold for cleaning bricks and roof tiles. (a) What is the molarity of this solution if 500mL of the concentrated solution contains 37 % (w/w) HCl and has a density of.8 g cm -3. Molar mass of HCl = 36.5 g mol Ans. Assume 00g of the solution Therefore in 00 g of the solution 37 g of HCl is present nhcl = 37 g 36.5 g mol =.0 mols From density (ρ) = mass(g) vol(ml). vol HCl = mass/ρ = 37 g /.8 g ml = 3.36 ml

7 Therefore MHCl = n (mols ) vol(dm 3 ) =.0 mol dm 3 = 32.2 mol dm-3 or 32.2 M (b) How would you prepare a 6 M HCl solution from the concentrated reagent? Assume preparation of 00 ml MHCl (init) vol HCl(init) = MHCl (final) vol HCl(final) 32.2 M (init) vol HCl(init) = 6M 00 ml MNH 3 (final) = 8.63 ml Dilute 8.63 ml of the concentrated HCl to 00 ml with distilled water. 0. Calculate the volume of 0.0 M NaOH that is required to neutralize: (a) 25.0 ml of 0.M HCl Ans. NaOH + HCl NaCl + H2O nnaoh = nhcl MNaOH x MHCl x Vol HCl 25.0 ml x 0. M HCl 0.0 M NaOH ml or dm 3

8 (b) 25.0 ml of 0. M H2SO4 Ans. 2NaOH + H2SO4 Na2SO4 + 2H2O nnaoh 2 = H2SO4 MNaOH x MH 2SO 4 x Vol H 2SO 4 x 2 M H2SO4 x Vol H2SO4 x 2 M NaOH 0. M x 25.0 ml x M 500 ml or 0.05 dm 3 (c) 25.0 ml of 0.M H3PO4 Ans. 3 NaOH + H3PO4 Na3PO4 + 3H2O nnaoh 3 = nh3po4 MNaOH x MH 3PO 4 x Vol H 3PO 4 x 3 M H3PO4 x Vol H3PO4 x 3 M NaOH 0. M x 25.0 ml x M 625 ml or dm 3

Steward Fall 08. Moles of atoms/ions in a substance. Number of atoms/ions in a substance. MgCl 2(aq) + 2 AgNO 3(aq) 2 AgCl (s) + Mg(NO 3 ) 2(aq)

Steward Fall 08. Moles of atoms/ions in a substance. Number of atoms/ions in a substance. MgCl 2(aq) + 2 AgNO 3(aq) 2 AgCl (s) + Mg(NO 3 ) 2(aq) Dealing with chemical stoichiometry Steward Fall 08 of Not including volumetric stoichiometry of Chapter 6.0x10 A 6.0x10 Mol/mol ratio from balanced equation B 6.0x10 6.0x10 s, Equations, and Moles: II