Question

A mixture 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?

Answer 1

The balanced chemical equation: 
N2 + 3H2 → 2NH3 

We see that 1mol N2 reacts with 3mol H2 to produce 2mol NH3 
But we produced 1.0mol NH3 
We rewrite the word equation to reflect this change by dividing all numbers by 2.
0.5Mol N2 reacts with 1.5mol H2 to produce 1.0mol NH3 
Original amounts: 
N2 = 1.0+0.5 = 1.5 mol N2 originally 
H2 = 1.0 + 1.5 = 2.5 mol H2 originally.

Answer 2

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,

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

Therefore, the amount of ${{\text{N}}_2}$  and ${{\text{H}}_2}$  are consumed until the reaction is stopped is 0.5 moles and 1.5 moles respectively.

Therefore, the amount of ${{\text{N}}_2}$  initially taken is,

$\begin{aligned}{\text{Amount of }}{{\text{N}}_2}&=\left({1.0 + 0.5}\right){\text{ mol}}\\&=1.5{\text{ mol}}\\\end{aligned}$

The amount of ${{\text{H}}_2}$  initially taken is,

$\begin{aligned}{\text{Amount of }}{{\text{H}}_2}&=\left({1.0 + 1.5}\right){\text{ mol}}\\&=2.5{\text{ mol}}\\\end{aligned}$

Learn more:

1. Balanced chemical equation: brainly.com/question/1405182

2. Determine how many moles of water produce: brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: N2, H2, NH3, 3H2, 2NH3, limiting reagent, nitrogen, hydrogen, ammonia, 0.5 mol, 1.0 mol, 1.5 mol, 2.5 mol, 1.0 mol of NH3.