Question

Two trials are run, using excess water. In the first trial, 7.8 g of Na2O2(s) (molar mass 78 g/mol) is mixed with 3.2 g of S(s). In the second trial, 7.8 g of Na2O2(s) is mixed with 6.4 g of S(s). The Na2O2(s) and S(s) react as completely as possible. Both trials yield the same amount of SO2(aq). Which of the following identifies the limiting reactant and the heat released, q, for the two trials at 298 K?
Limiting Reactant q
A. S 30. kJ
B. S 61 kJ
C. Na2O2 30. kJ
D. Na2S2 61 kJ

Answer 1

Answer:

The answer is "Option C".

Explanation:

Given equation:

$2Na_20_2 (s)+S(s)+2H_2O \longrightarrow 4NaOH(aq)+SO_2(aq)$

$\to \Delta H^{\circ}_{rxn} (298\ K) = -610 \frac{kJ}{mol}$

$\to Na_2O_2 \ Mass = 7.8 \ g\\\\ \to Na_2O_2 \ Molar \ mass = 78 \frac{g}{mol}$

$Na_2O_2$ Has been the reactant which is limited since the two experiments are equal to$Na_2O_2$ for relationship between stress amounts.

$Na_2O_2, n =\frac{Mass of Na_2O_2}{Molar mass of Na_2 O_2}=\frac{7.8 \ g}{78 \frac{g}{mol}} =0.1 \ mol \\\\q=\Delta H^{\circ}_{rxn} \times n = \frac{ -610 \ kJ}{ 2 \ mol \ Na_2 O_2} \times 0.1 \ mol \ Na_2O_2= 30.5 \ KJ$

Limiting reactant =$Na_2O_2$

$q=30.5 \ kJ \approx 30 \ kJ$