Chemistry, 03.07.2019 16:30, campbell387

Amixture of n2(g) and h2(g) reacts in a closed container to form ammonia, nh3(g). the reaction ceases before either reactant has been totally consumed. at this stage 1.0 mol n2, 1.0 mol h2, and 1.0 mol nh3 are present. part a how many moles of n2 and h2 were present originally?

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The amount of ${{\text{N}}_2}$ initially taken is $\boxed{{\text{1}}{\text{.5 mol}}}$ and the amount of ${{\text{H}}_2}$ initially taken is $\boxed{{\text{2}}{\text{.5 mol}}}$ .

Further explanation:

Stoichiometry of a reaction is used to determine the amount of species present in the reaction by the relationship between the reactants and products. It can be used to determine the moles of a chemical species when the moles of other chemical species present in the reaction is given.

Consider the general reaction,

${\text{A}}+2{\text{B}}\to3{\text{C}}$

Here,

A and B are reactants.

C is the product.

One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

The balanced chemical equation for the formation of ${\text{N}}{{\text{H}}_3}$ is as follows:

${{\text{N}}_2}\left(g\right)+3{{\text{H}}_2}\left(g\right)\to2{\text{N}}{{\text{H}}_3}\left(g\right)$

The balanced chemical equation shows that 1 mole of ${{\text{N}}_2}$ and 3 moles of ${{\text{H}}_2}$ reacts to form 2 moles of ${\mathbf{N}}{{\mathbf{H}}_{\mathbf{3}}}$ .

In the question, reaction started with the unknown quantity of reactant ${{\text{N}}_2}$ and ${{\text{H}}_2}$ and stopped when 1.0 mol ammonia was produced. Also, reactants left in the reaction mixture are 1.0 mol of ${{\text{H}}_2}$ and 1.0 mol of ${{\text{N}}_2}$ .

According to the balance reaction 2 moles of ammonia is produced by the 3 moles of ${{\text{H}}_2}$ . Thus amount of hydrogen molecule required to produce 1 mole of ammonia is,

$\begin{gathered}{\text{Moles of }}{{\text{H}}_2}=\frac{{3{\text{ mol }}{{\text{H}}_2}}}{{2{\text{ mol N}}{{\text{H}}_3}}}\times1{\text{ mol N}}{{\text{H}}_3}\\=1.5{\text{ mol }}{{\text{H}}_{\text{2}}}\\\end{gathered}$

According to the balance reaction 2 moles of ammonia is produced by the 1 mole of ${{\text{N}}_2}$ . Thus the amount of ${{\text{N}}_2}$ molecule required to produce 1 mole of ammonia is,

$\begin{gathered}{\text{Moles of }}{{\text{N}}_2}=\frac{{1{\text{ mol }}{{\text{N}}_2}}}{{2{\text{ mol N}}{{\text{H}}_3}}}\times1{\text{ mol N}}{{\text{H}}_3}\\=0.5{\text{ mol }}{{\text{N}}_2}\\\end{gathered}$