Question

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation.
${N_2}\left( g \right) + {H_2}\left( g \right)\xrightarrow{{}}2N{H_3}\left( g \right)$
i.Calculate the mass of ammonia produced if $2 \times {10^3}\,g$ Dinitrogen reacts with $1 \times {10^3}g$ of hydrogen.
ii.Will any of the two reactants remain unreacted?
iii.If yes, which one and what would be its mass?

Answer 1

Hint:we know that a mole ratio is a ratio between the numbers of moles of any two species involved in a chemical reaction.
Example,
In the reaction${\text{2}}{{\text{H}}_{\text{2}}}{\text{ + }}{{\text{O}}_{\text{2}}}\xrightarrow{{}}{\text{2}}{{\text{H}}_{\text{2}}}{\text{O}}$, the mole ratio can be written as $\dfrac{{{\text{2}}\,{\text{mol}}\,{{\text{H}}_{\text{2}}}}}{{{\text{1}}\,{\text{mol}}\,{{\text{O}}_{\text{2}}}}}.$

Complete step by step answer:
The given equation is,
${N_2}\left( g \right) + 3{H_2}\left( g \right)\xrightarrow{{}}2N{H_3}\left( g \right)$
Twenty eight gram of nitrogen reacts with six gram of hydrogen to give ammonia.
The total mass of ammonia$ = 2 \times 14 + 2\left( 3 \right) = 34$
One gram of nitrogen reacts with one gram of hydrogen to give $34/28 \times 1\,g$ ammonia.
When $2 \times {10^3}\,g$ of Dinitrogen reacts with $1 \times {10^3}g$ of hydrogen to give,
$34/28 \times 2 \times {10^3}\,g = 2428.57\,g$
The mass of ammonia produced if $2 \times {10^3}\,g$ Dinitrogen reacts with $1 \times {10^3}g$ of hydrogen is $2428.57\,g$.
Hence nitrogen is the limiting agent.
Dinitrogen is the limiting reagent and hydrogen is the excess reagent. Hence, dihydrogen will remain unreacted.
Amount of hydrogen that remains unreacted.
Twenty eight gram of nitrogen reacts with six gram of hydrogen to give ammonia. Hence $2 \times {10^3}\,g$ require.$ = \dfrac{6}{{28}} \times 2 \times {10^3} = 428.5\,g$
Amount of hydrogen that remains unreacted.$ = 1 \times {10^3}\,g - 428.5\,g = 571.5\,g$

Additional information:
Example:
The number of grams of hydrochloric acid in the following reaction can be calculated using the mole ratio,
$2{C_{12}}{H_4}C{l_6} + 23\,{O_2} + 2{H_2}O\xrightarrow{{}}24\,C{O_2} + 12HCl$
Given,
The number of moles of water is ${\text{15}}{\text{.2}}\,{\text{mol}}{\text{.}}$
The molecular mass of $HCl$ is $36.46\,g/mol$.
In the mole ratio, the coefficients of the balanced equation are used. Therefore the mole ratio is $\dfrac{{12\,mol\,HCl}}{{2\,mol\,{H_2}O}}.$
The number of moles can be calculated as,
$15.2\,mol\,{H_2}O\,\left( {\dfrac{{12\,mol\,HCl}}{{2\,mol\,{H_2}O}}} \right) \times \dfrac{{36.46\,g}}{{1\,mol}} = 3325\,g\,HCl$
The number of grams of ${\text{HCl}}$ is $3325\,g.$

Note:
The limiting reactant of the reaction is the reactant that is completely used during the reaction. Using the mole ratio and starting amounts of the reactants limiting reactant can be determined.
Example:
Consider a reaction starting with $30\,g{\text{ }}CaC{O_3}\,\& \,11\,g\,HCl.$The values in grams have to be converted to moles by dividing with their molecular weights. According to the mole ratio $0.3\,g$ of calcium carbonate requires $0.6\,g$ of $HCl$ completely. Therefore $HCl$ is the limiting agent.