If 23.7 g of Al(OH)3(s) are mixed with 29.5 g of H2SO4(s) and the reaction is run, answer the following questions:

1 answer:

3 0

Aluminum hydroxide $\text{Al}(\text{OH})_3$ can behave as a base and neutralize sulfuric acid $\text{H}_2\text{SO}_4$ as in the following equation:

$2\;\text{Al}(\text{OH})_3 \; (s) + 3\; \text{H}_2\text{SO}_4 \; (aq) \to \text{Al}_2(\text{SO}_4)_3 \; (aq) + 6 \; \text{H}_2\text{O} \; (l)$ (Balanced)

(a)

$n = m/M$ . Thus the ratio between the number of moles of the two reactants available:

$n(\text{Al}(\text{OH})_3, \text{supplied}) / n(\text{H}_2\text{SO}_4, \text{supplied})\\= [m(\text{Al}(\text{OH})_3)/ M(\text{Al}(\text{OH})_3)] / [n(\text{H}_2\text{SO}_4) / M(\text{H}_2\text{SO}_4)]\\= [23.7 / (26.98 + 3 \times(16.00 + 1.008))]/[29.5 / (2 \times 1.008 + 32.07 + 4 \times 16.00)]\\\approx 1.01$

The value of this ratio required to lead to a complete reaction is derived from coefficients found in the balanced equation:

$n(\text{Al}(\text{OH})_3, \text{theoretical}) / n(\text{H}_2\text{SO}_4, \text{theoretical}) = 2/3 \approx 0.667$

The ratio for the complete reaction is smaller than that of the reactants available, indicating that the species represented on the numerator, $\text{Al}(\text{OH})_3$ , is in excess while the one on the denominator, $\text{H}_2\text{SO}_4$ , serves as the limiting reagent.

(b)

The quantity of water produced is dependent on the amount of limiting reactants available. $29.5 / (2 \times 1.008 + 32.07 + 4 \times 16.00) = 0.301 \; \text{mol}$ of sulfuric acid is supplied in this reaction as the limiting reagent. $6$ moles of water molecules are produced for every $3$ moles of sulfuric acid consumed. The reaction would thus give rise to $0.301 \; \text{mol} \times 6/3 = 0.602 \; \text{mol}$ of water molecules, which have a mass of $0.602 \times (2 \times 1.008 + 16.00) = 10.8 \; \text{g}$ .

(c)

$\text{Percentage Yield}\\= \text{Actual Yield} / \text{Theoretical Yield} \times 100 \; \%\\= 2.21 / 10.8 \times 100 \; \%\\= 20.4 \; \%$

(d)

The quantity of $\text{Al}(\text{OH})_3$ , the reactant in excess, is dependent on the number of moles of this species consumed in the reaction and thus the quantity of the limiting reagent available. The consumption of every $3$ moles of sulfuric acid, the limiting reagent, removes $2$ moles of aluminum hydroxide $\text{Al}(\text{OH})_3$ from the solution. $0.301 \; \text{mol}$ of sulfuric acid is initially available as previously stated such that $0.301 \; \text{mol} \times 2/3 = 0.201 \; \text{mol}$ , or $0.201 \times (26.98 + 3 \time (16.00 + 1.008)) = 15.7 \; \text{g}$ , of $\text{Al}(\text{OH})_3$ would be eventually consumed.