# Hydrogen and oxygen react chemically to form water. How much water would form if 4.8 grams of hydrogen reacted with 38.4 grams of oxygen?

Sep 5, 2016

Approx. $36 \cdot g$

#### Explanation:

We need a stoichiometric equation for water synthesis:

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$

Clearly, dihydrogen must be present in a 2:1 molar ratio with respect to dioxygen.

$\text{Moles of dihydrogen}$ $=$ $\frac{4.8 \cdot g}{2.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.39 \cdot m o l$

$\text{Moles of dioxygen}$ $=$ $\frac{38.4 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.2 \cdot m o l$

Given these molar quantities (they are approx. 2:1) the gases are present in stoichiometric proportion. $2$ $m o l$ water are going to be produced.

$2 \cdot m o l \times 18.00 \cdot g \cdot m o {l}^{-} 1$ $=$ $36 \cdot g$

Do you think that this would be an exothermic reaction; i.e. do you think that energy would be released by the reaction?