Question

1. Calculate the amount in moles and the mass in grams of KOH needed to convert 1.50 g of
aluminum to KAl(OH) . Write first the balanced equation.
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2. Calculate the volume per mL of 9.0M H2SO4 that you would need to convert 0.070 moles of
KAl(OH)4 to K2SO4 and Al2(SO4)3 according to the following equation:
2 KAl(OH)4 + 4 H2SO4 → K2SO4 + Al2(SO4)3 + 8 H2O

Answer 1

$1. \ KOH+Al + 3H_2O→K(Al)OH_4​ +\frac{3}{2}H_2$

1 mol of KOH reacts with 1 mol of Al

56g of KOH reacts with 27g of Al

xg of KOH reacts with 1.50g of Al

x = 56/27 × 1.50 = 3.11g

3.11g of KOH = 3.11/56 mol = 0.056 mol

Therefore, 3.11g (0.056 mol) of KOH converts 1.50g of Al in the reaction.

2. According to the reaction,

2 moles of $K(Al)OH_4$ reacts with 4 moles of $H_2SO_4$

$\therefore$ 0.070 moles of $K(Al)OH_4$

reacts with 0.140 (0.070×2) moles of $H_2​SO_4​$

Concentration of $H_2​SO_4​$ = 9M.

To get the volume needed,

9M = 0.140/Vol

Vol = 0.140/9 = 0.0156L = 15.6mL.

Therefore, 15.6mL of $H_2SO_4$ is needed.