# 4 gms of hydrogen are ignited with 4 gms of oxygen. the weight of water formed ?

May 22, 2017

$\text{Mass} \cong 4.5 \cdot g$

#### Explanation:

We need (i), a stoichiometric equation,

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

We have $\frac{4 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1} = \frac{1}{8} \cdot m o l$ $\text{dioxygen gas}$.

And $\frac{4 \cdot g}{2.0 \cdot g \cdot m o {l}^{-} 1} = 2 \cdot m o l$ $\text{dihydrogen gas}$.

Clearly, the $\text{dioxygen gas}$ is present in DEFICIENCY, and ${O}_{2}$ is the limiting reagent.

Given the equation, we can make at most $\frac{1}{4} \cdot m o l$ of water, i.e. a mass of 1/4*cancel(mol)xx18.01*g*cancel(mol^-1)=??*g