Chemistry, 16.01.2020 10:31 balletbella0531

Nitrogen and hydrogen combine at high temperature, in the presence of a catalyst, to produce ammonia. n2(g)+3h2(g)=2nh3(g). assume 0.280 mol of n2 and 0.884 mol of h2 are present initially.. after complete reaction, how many moles of ammonia are . how many moles of h2 . how many moles of n2, what is the limiting or hydrogen)

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Answer from: domp40

The balanced chemical reaction is:

N2(g)+3H2(g)=2NH3(g)

We are given the amount of the reactants to be used. These are the starting values to be used in the calculations. In this case, nitrogen gas is the limiting reactant.

0.280 mol N2 ( 2 mol NH3 / 1 mol N2 ) = 1.6 mol NH3

0.884 - 0.280 mol N2 ( 3 mol H2 / 1 mol N2 ) = 0.044 mol H2 excess

0 mol N2 remains all are consumed by the reaction.

Answer from: dashaunpeele

Number of moles of ammonia produced on completion of reaction is $\boxed{{\text{0}}{\text{.560 moles}}}$

Number of moles of ${{\text{H}}_2}$ remain on completion of reaction is $\boxed{{\text{0}}{\text{.044 moles}}}$

Number of moles of ${{\text{N}}_2}$ remain on completion of reaction is $\boxed{{\text{0 moles}}}$

The limiting reagent of the reaction is $\boxed{{{\text{N}}_2}{\text{ molecule}}}$

Further Explanation:

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction. The amount of product depends on the amount of limiting reagent since the product formation is not possible in the absence of it.

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

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

From the balanced chemical reaction, the reaction stoichiometry between ${{\text{N}}_2}$ and ${{\text{H}}_2}$ is as follows:

$1{\text{mol}}{{\text{N}}_2}:3{\text{mol}}{{\text{H}}_2}$

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 ${\text{N}}{{\text{H}}_3}$ but the reaction mixture has 0.280 moles of ${{\text{N}}_2}$ and 0.884 moles of ${{\text{H}}_2}$ .

Calculate the moles of ${{\text{N}}_2}$ that react with 0.884 moles of ${{\text{H}}_2}$ as follows:

$\begin{aligned}{\text{Amount of }}{{\text{N}}_{\text{2}}}\left({{\text{mol}}}\right)&=\left({{\text{0}}{\text{.884 mol }}{{\text{H}}_2}}\right)\left({\frac{{1{\text{ mol }}{{\text{N}}_2}}}{{3{\text{ mol }}{{\text{H}}_2}}}}\right)\\&=0.295{\text{ mol }}{{\text{N}}_2}\\\end{aligned}$

The calculations concluded that 0.295 moles of ${{\text{N}}_2}$ is required for 0.884 moles of ${{\text{H}}_2}$ but we have only 0.280 moles of ${{\text{N}}_2}$ in reaction mixture. Hence, ${{\mathbf{N}}_{\mathbf{2}}}$ is present in limited quantity and is a limiting reagent.

According to the balanced chemical equation 1 mole of ${{\text{N}}_2}$ produces 2 moles of ${\text{N}}{{\text{H}}_3}$ , therefore, the moles of ${\text{N}}{{\text{H}}_3}$ produced by 0.280 moles of ${{\text{N}}_2}$ can be calculated as follows:

$\begin{aligned}{\text{Amount of N}}{{\text{H}}_3}\left({{\text{mol}}}\right)&=\left( {{\text{0}}{\text{.280 mol }}{{\text{N}}_2}}\right)\left( {\frac{{2{\text{ mol N}}{{\text{H}}_3}}}{{1{\text{ mol }}{{\text{N}}_2}}}}\right)\\&={\mathbf{0}}{\mathbf{.560 mol N}}{{\mathbf{H}}_{\mathbf{3}}}\\\end{aligned}$

According to the balanced chemical equation 1 mole of ${{\text{N}}_2}$ reacts with 3 moles of ${{\text{H}}_{\text{2}}}$ , therefore, the number of moles of ${{\text{H}}_{\text{2}}}$ reacted by 0.280 moles of ${{\text{N}}_2}$ can be calculated as follows:

$\begin{aligned}{\text{Amount of }}{{\text{H}}_2}\left( {{\text{mol}}} \right)&=\left({{\text{0}}{\text{.280 mol }}{{\text{N}}_2}}\right)\left({\frac{{3{\text{ mol }}{{\text{H}}_2}}}{{1{\text{ mol }}{{\text{N}}_2}}}}\right)\\&={\mathbf{0}}{\mathbf{.840 mol }}{{\mathbf{H}}_{\mathbf{2}}}\\\end{aligned}$

When reaction is completed, 0.840 moles of ${{\text{H}}_2}$ is consumed out of 0.884 moles of ${{\text{H}}_2}$ . Therefore, the number of moles of ${{\text{H}}_2}$ remained after the completion is calculated as follows:

$\begin{aligned}{\text{Amount of }}{{\text{H}}_2}{\text{ remains}}&=\left({0.884 - 0.840} \right){\text{ mol of }}{{\text{H}}_2}\\&={\mathbf{0}}{\mathbf{.044 mol of }}{{\mathbf{H}}_{\mathbf{2}}}\\\end{aligned}$

Since ${{\text{N}}_2}$ molecule is a limiting reagent thus it is consumed completely on completion of reaction. Therefore, the amount of ${{\mathbf{N}}_{\mathbf{2}}}$ remains after completion of reaction is zero moles.

Learn more:

1. Balanced chemical equation:

2. Determine how many moles of water produce:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: N2, H2, NH3, 3H2, 2NH3, limiting reagent, nitrogen, hydrogen, ammonia, 0.280 mol of N2, 0.884 mol of H2, 0.044 mol H2.

Answer from: Quest

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Answer from: Quest

It shows that the experiment is designed properly.