Answer:
The answer is "Option C".
Explanation:
Given equation:
[tex]2Na_20_2 (s)+S(s)+2H_2O \longrightarrow 4NaOH(aq)+SO_2(aq)[/tex]
[tex]\to \Delta H^{\circ}_{rxn} (298\ K) = -610 \frac{kJ}{mol}[/tex]
[tex]\to Na_2O_2 \ Mass = 7.8 \ g\\\\ \to Na_2O_2 \ Molar \ mass = 78 \frac{g}{mol}[/tex]
[tex]Na_2O_2[/tex] Has been the reactant which is limited since the two experiments are equal to[tex]Na_2O_2[/tex] for relationship between stress amounts.
[tex]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\\\\[/tex]
Limiting reactant =[tex]Na_2O_2[/tex]
[tex]q=30.5 \ kJ \approx 30 \ kJ[/tex]
