How much heat will be released when 12.0 g of H_2 reacts with 76.0 g of O_2 according to the following equation? 2H_2 + O_2 -> 2H_2O DeltaH= -571.6 kJ?

Nov 19, 2016

$\text{Dioxygen gas}$ is the limiting reagent. Over $1.3 \times {10}^{3} \cdot k J$ are evolved.

Explanation:

$2 {H}_{2} \left(g\right) + {O}_{2} \left(g\right) \rightarrow 2 {H}_{2} O \left(l\right)$ $\Delta H = - 576.6 \cdot k J \cdot m o {l}^{-} 1.$

$\text{Moles of dihydrogen} = \frac{12.0 \cdot g}{2.016 \cdot g \cdot m o {l}^{-} 1} = 5.95 \cdot m o l$.

$\text{Moles of dioxygen} = \frac{76.0 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1} = 2.38 \cdot m o l$.

Given the stoichiometry, clearly, there is a insufficient molar quantity of dioxygen for complete combustion. At most $4.75 \cdot m o l$ dihydrogen can react (i.e. 2xx2.38*mol) according to the given equation.

And thus energy released can based on the molar quantity of $\text{dioxygen gas}$ $=$ 2.38*molxx-576.6*kJ*mol^-1=??kJ