Respuesta :
The amount of [tex]{{\text{N}}_2}[/tex] initially taken is [tex]\boxed{{\text{1}}{\text{.5 mol}}}[/tex] and the amount of [tex]{{\text{H}}_2}[/tex] initially taken is [tex]\boxed{{\text{2}}{\text{.5 mol}}}[/tex].
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,
[tex]{\text{A}}+2{\text{B}}\to3{\text{C}}[/tex]
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 [tex]{\text{N}}{{\text{H}}_3}[/tex] is as follows:
[tex]{{\text{N}}_2}\left(g\right)+3{{\text{H}}_2}\left(g\right)\to2{\text{N}}{{\text{H}}_3}\left(g\right)[/tex]
The balanced chemical equation shows that 1 mole of [tex]{{\text{N}}_2}[/tex] and 3 moles of [tex]{{\text{H}}_2}[/tex] reacts to form 2 moles of [tex]{\mathbf{N}}{{\mathbf{H}}_{\mathbf{3}}}[/tex] .
In the question, reaction started with the unknown quantity of reactant [tex]{{\text{N}}_2}[/tex] and [tex]{{\text{H}}_2}[/tex] and stopped when 1.0 mol ammonia was produced. Also, reactants left in the reaction mixture are 1.0 mol of [tex]{{\text{H}}_2}[/tex] and 1.0 mol of [tex]{{\text{N}}_2}[/tex] .
According to the balance reaction 2 moles of ammonia is produced by the 3 moles of [tex]{{\text{H}}_2}[/tex] . Thus amount of hydrogen molecule required to produce 1 mole of ammonia is,
[tex]\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}[/tex]
According to the balance reaction 2 moles of ammonia is produced by the 1 mole of [tex]{{\text{N}}_2}[/tex] . Thus the amount of [tex]{{\text{N}}_2}[/tex] molecule required to produce 1 mole of ammonia is,
[tex]\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}[/tex]
Therefore, the amount of [tex]{{\text{N}}_2}[/tex] and [tex]{{\text{H}}_2}[/tex] are consumed until the reaction is stopped is 0.5 moles and 1.5 moles respectively.
Therefore, the amount of [tex]{{\text{N}}_2}[/tex] initially taken is,
[tex]\begin{aligned}{\text{Amount of }}{{\text{N}}_2}&=\left({1.0 + 0.5}\right){\text{ mol}}\\&=1.5{\text{ mol}}\\\end{aligned}[/tex]
The amount of [tex]{{\text{H}}_2}[/tex] initially taken is,
[tex]\begin{aligned}{\text{Amount of }}{{\text{H}}_2}&=\left({1.0 + 1.5}\right){\text{ mol}}\\&=2.5{\text{ mol}}\\\end{aligned}[/tex]
Learn more:
1. Balanced chemical equation: https://brainly.com/question/1405182
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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.
The number of moles of N2 and H2 that were originally present are;
N2 = 1.5 mol
N2 = 1.5 molH2 = 2.5 mol
- We are told that N2(g) reacts with H2(g) to form ammonia NH3(g).
This reaction when balanced is;
N2(g) + 3H2(g) = 2NH3(g)
- Now, we are told that the reaction ceases before the reactants are totally consumed and that there was 1 mole of ammonia present. This means we have to simplify our balanced equation so that the number of moles attached to Ammonia(NH3) can be 1.
Thus,divide each of the number of moles in the balanced equation by 2 to get;
0.5N2(g) + 1.5H2(g) = NH3(g)
- We are told that the number of moles of N2 and H2 that were present at the stage after the reaction had ceased was;
1 mol of N2 and 1 mol of H2.
Thus;
Number of moles of N2 originally present = 1 + 0.5 = 1.5 moles
Number of moles of H2 originally present = 1 + 1.5 = 2.5 moles
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