Question

What mass of ${H_2}O$ is formed when ${H_2}$ reacts with 384 g of $\;{0_2}$ ?

Answer 1

The mole concept is very significant and useful in chemistry. It is actually the base of stoichiometry and it provides the best option to express the amounts of reactants as well as products that are consumed and formed during a chemical reaction.

Complete step by step answer:
To calculate the number of moles, we generally use the following formula:
$Number{\text{ }}of{\text{ }}moles = \dfrac{{Given{\text{ }}mass{\text{ }}}}{{{\text{ M}}olecular{\text{ }}mass{\text{ }}of{\text{ }}the{\text{ }}given{\text{ }}species}}$
Alternatively, we can also depict the number of moles from the balanced chemical equation. The coefficients actually represent the number of moles which react, not just only molecules.
In the question we are given that ${H_2}$ reacts with $\;{0_2}$ to form ${H_2}O$ . The balanced chemical equation can be written as follows:
$2{H_2} + {O_2} \to 2{H_2}O$
This chemical equation indicates that 2 moles of ${H_2}$ reacts with 1 mole of $\;{0_2}$ to form 2 mole of ${H_2}O$ . That means the mole ratio between oxygen and water is $1:2$ .
In the question we are given a mass of $\;{0_2}$ as 384 g. And we know molar mass can be found out by adding the relative atomic masses of each element present in the compound. We know that the atomic mass of oxygen is 16 so this indicates that the molar mass of $\;{0_2}$ will be 32 $gmo{l^{ - 1}}$ . We can substitute these values to calculate the number of moles of $\;{0_2}$ as follows:
$Number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}{O_2} = \dfrac{{384g{\text{ }}}}{{{\text{ }}32gmo{l^{ - 1}}}} = 12mol{\text{ }}of{\text{ }}{O_2}$
We know the mole ratio between oxygen and water is $1:2$ . So here 12 moles of oxygen are used, that means $12 \times 2 = 24\;mol\;$ of water is produced. We can calculate the molar mass of ${H_2}O$ as stated below:
$Molar{\text{ }}mass{\text{ }}of{\text{ }}{H_2}O = \;(2 \times 1) + (1 \times 16) = 18\;gmo{l^{ - 1}}$
Substituting the values in number of moles formula, we can calculate the mass of ${H_2}O$ as shown below:

$$24 = \dfrac{{Mass{\text{ }}}}{{{\text{ }}18}} \\\ Mass{\text{ }}of{\text{ }}{H_2}O = 24 \times 18 = 432g \\\$$

Hence, 432 g of ${H_2}O$ is formed when ${H_2}$ reacts with 384 g of $\;{0_2}$ .

Note: A balanced chemical equation simply obeys the law of conservation of mass. Balancing the chemical equations is a significant guiding principle in chemistry. A balanced chemical equation helps you to predict the amount of reactants required and the amount of products formed.