Question

if 50.0 g of h₂ and 158.5 g of o₂ react, how many moles of h₂o can be produced in the reaction below? 2 h₂(g) + o₂(g) → 2 h₂o(g)

Explanation:

Step1: Calculate moles of $H_2$

The molar mass of $H_2$ is $M_{H_2}=2\times1\ g/mol = 2\ g/mol$. Using the formula $n=\frac{m}{M}$, where $m$ is mass and $M$ is molar - mass, for $H_2$: $n_{H_2}=\frac{50.0\ g}{2\ g/mol}=25.0\ mol$.

Step2: Calculate moles of $O_2$

The molar mass of $O_2$ is $M_{O_2}=2\times16\ g/mol = 32\ g/mol$. Using the formula $n=\frac{m}{M}$, for $O_2$: $n_{O_2}=\frac{158.5\ g}{32\ g/mol}=4.953\ mol$.

Step3: Determine the limiting reactant

From the balanced chemical equation $2H_2(g)+O_2(g)
ightarrow2H_2O(g)$, the mole - ratio of $H_2$ to $O_2$ is $2:1$. For $4.953\ mol$ of $O_2$, the amount of $H_2$ required is $n_{H_2\ required}=2\times n_{O_2}=2\times4.953\ mol = 9.906\ mol$. Since we have $25.0\ mol$ of $H_2$ and only $9.906\ mol$ are required to react with all of the $O_2$, $O_2$ is the limiting reactant.

Step4: Calculate moles of $H_2O$ produced

The mole - ratio of $O_2$ to $H_2O$ is $1:2$. Using the amount of the limiting reactant ($O_2$), the number of moles of $H_2O$ produced is $n_{H_2O}=2\times n_{O_2}$. Since $n_{O_2}=4.953\ mol$, $n_{H_2O}=2\times4.953\ mol = 9.906\ mol$.

Answer:

$9.906\ mol$