# Question 77fa3

Jan 16, 2018

$0.75$ moles of ${H}_{2} O$ and $0.375$ moles of ${O}_{2}$

#### Explanation:

According to the equation, $2 {H}_{2} {O}_{2} \rightarrow 2 {H}_{2} O + {O}_{2}$

2 moles of ${H}_{2} {O}_{2}$ produces 2 moles of ${H}_{2} O$ and 1 mole of ${O}_{2}$

Therefore, relating, $0.75$ moles of ${H}_{2} {O}_{2}$ would produce $0.75$ moles of ${H}_{2} O$ and $0.375$ $\left(a s \frac{0.75}{2} = 0.375\right)$ moles of ${O}_{2}$

Cheers,
-Sahar;)

Jan 16, 2018

$n \left({H}_{2} O\right) = 0.75 \text{ mol}$,
$n \left({O}_{2}\right) = 0.375 \text{ mol}$.

#### Explanation:

First, let's find the stoichiometric ratio between the two reactants and the product.

We may derive a couple of conversion factors from the balanced chemical equation
$2 {H}_{2} {O}_{2} \left(l\right) \to 2 {H}_{2} O \left(l\right) + {O}_{2} \left(g\right)$.

To simplify our calculations, we will try to put the substance with the known molar quantity (${H}_{2} {O}_{2}$ in our case) on the denominator.

From coefficients in the equation we find ratios
$\left(2 \text{ mol "H_2O)/(2" mol } {H}_{2} {O}_{2}\right)$ and $\left(1 \text{ mol "O_2)/(2" mol } {H}_{2} {O}_{2}\right)$

The question states that the number of moles of hydrogen peroxide ${H}_{2} {O}_{2}$ is $0.75$ moles. We can use that value in our calculation since ${H}_{2} {O}_{2}$ is the only reactant in this reaction.

$n \left({H}_{2} O\right)$
=0.75" mol "H_2O_2 * (2" mol "H_2O)/(2" mol " H_2O_2)
$= 0.75 \text{ mol } {H}_{2} O$

$n \left({O}_{2}\right)$
=0.75" mol " H_2O_2 *(1" mol "O_2)/(2" mol " H_2O_2)#
$= 0.375 \text{ mol } {H}_{2} O$

(You are supposed to use only the number of moles of the limiting reagent when doing stoichiometrical calculations, so be careful dealing with reactions involving more than one reactants.)

Check your answers using dimensional analysis by making sure that unit for the answer is correct (for example, expect * $\text{mol } {H}_{2} O$ for number of moles of water;) unrelated units shall cancel out in the calculation.*

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