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4 Aqueous Reactions Solutions to Exercises (d) Plan. The amount of the limiting reactant (KOH) determines amount of product, in this case Ni(OH)₂. Solve. 0.0200 mol 1 mol Ni(OH)₂ = 0.927 Ni(OH)₂ 2 mol 1 mol Ni(OH)₂ (e) Plan/Solve. Limiting reactant: OH-: no excess OH- remains in solution. Excess reactant: Ni²⁺: M Ni²⁺ remaining = mol Ni²⁺ remaining/L solution 0.0300 mol Ni²⁺ initial - 0.0100 mol Ni²⁺ reacted = 0.0200 mol Ni²⁺ remaining 0.0200 mol /0.3000 L = 0.0667 M (aq) Spectators: These ions do not react, so the only change in their concentration is dilution. The final volume of the solution is 0.3000 L. M₂ = M₁V₁/V₂: 0.200 X 0.1000 L/0.3000 L = 0.0667 M (aq) 0.150 M X 0.2000 L/0.3000 L = 0.100 M (aq) 4.88 (a) Sr(NO₃)₂(aq) + (b) Determine the limiting reactant, then the identity and concentration of ions remaining in solution. Assume that the produced by the reaction does not increase the total solution volume. 15.0 g Sr(OH)₂ X 1 mol Sr(OH)₂ = 0.1233 = 0.123 mol Sr(OH)₂ mol = 2(0.1233) mol Sr(OH)₂ = 0.2466 = 0.247 mol 0.200 M X 0.0550 L = 0.0110 mol Two mol react with one mol Sr(OH)₂, so is the limiting reactant. No excess remains in solution. The remaining ions are OH- (excess reactant), and (spectators). 0.2466 mol OH- initial - 0.0110 mol react = 0.2356 = 0.236 mol remain 0.2356 mol OH-/0.0550 L soln = 4.28 M (aq) 0.123 mol /0.0550 L soln = 2.24 M (aq) 0.0110 mol /0.0550 L = 0.200 M (aq) (c) The resulting solution is basic because of the large excess of (aq). 4.89 Analyze. Given: mass impure Mg(OH)₂; M and vol excess HCl; M and vol NaOH. Find: mass % Mg(OH)₂ in sample. Plan/Solve. Write balanced equations. Mg(OH)₂(s) + HCl(aq) + NaOH(aq) + Calculate total moles HCI = M L 0.2050 mol x0.1000 L = 0.02050 mol total 1L soln 98