# When 1.57 mol O_2 reacts with H_2 to form H_2O, how many moles of H_2 are consumed in the process?

Dec 9, 2016

Look at the stoichiometry, the chemical equivalence, of the reaction:

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

#### Explanation:

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$, or I could simply double the equation,

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

Now, clearly, for each mole/equiv of dioxygen, 2 moles/equivs dihydrogen react. If there are $1.57 \cdot m o l$ ${O}_{2}$, stoichiometric equivalence requires $2 \times 1.57 \cdot m o l$ dihydrogen. (i) How many moles of water will be generated; and (ii) what is the mass of the water?