Respuesta :
The complete balanced reaction for this is:
N2 + 3H2 --> 2NH3
First we convert the given masses into number of moles.
Molar mass of N2 = 28 g/mol
Molar mass of H2 = 2 g/mol
Therefore,
moles N2 = 14.01 / 28 = 0.5 moles
moles H2 = 3.02 / 2 = 1.51 moles
Then we find which has lower moles/coefficient ratio:
N2 = 0.5 / 1 = 0.5
H2 = 1.51 / 3 = 0.503
Since N2 has lower moles/coefficient ratio, therefore it is the limiting reactant.
So total moles of ammonia formed is:
moles NH3 = 0.5 moles N2 * (2 moles NH3 / 1 mole N2) = 1 mole NH3
The molar mass of NH3 is 17.031 g/mol, hence:
mass NH3 = 1 mole * 17.031 g/mol
mass NH3 = 17.031 grams
The mass of ammonia that is produced by the reaction between [tex]{{\text{N}}_2}[/tex] and [tex]{{\text{H}}_2}[/tex] is [tex]\boxed{{\mathbf{34}}{\mathbf{.06 g}}}[/tex].
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 balanced chemical equation that represents 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) \to 2{\text{N}{{\text{H}}_3}\left( g \right)[/tex]
From the balanced chemical reaction, the reaction stoichiometry between [tex]{{\text{N}}_2}[/tex] and [tex]{{\text{H}}_2}[/tex] is as follows:
[tex]1{\text{ mol }}{{\text{N}}_2}:3{\text{ mol }}{{\text{H}}_2}[/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]{\text{N}}{{\text{H}}_3}[/tex].
First, we have to calculate the actual number of moles of [tex]{{\text{N}}_2}[/tex] and [tex]{{\text{H}}_2}[/tex] present in the mixture.
The mass of [tex]{{\text{N}}_2}[/tex] present in the reaction mixture is 14.01 g and mass of [tex]{{\text{H}}_2}[/tex] present in the mixture is 3.02 g.
The number of moles of [tex]{{\text{N}}_2}[/tex] present in the reaction mixture is calculated as,
[tex]\begin{aligned}{\text{Moles of }}{{\text{N}}_2} &=\frac{{{\text{Given mass of }}{{\text{N}}_2}}}{{{\text{Molar mass of }}{{\text{N}}_2}}}\\&=\frac{{14.01{\text{ g}}}}{{14.01{\text{ g/mol}}}}\\&=1.0{\text{ mol}}\\\end{aligned}[/tex]
The number of moles of [tex]{{\text{H}}_2}[/tex] present in the reaction mixture is calculated as,
[tex]\begin{aligned}{\text{Moles of }}{{\text{H}}_2} &=\frac{{{\text{Given mass of }}{{\text{H}}_2}}}{{{\text{Molar mass of }}{{\text{H}}_2}}}\\&=\frac{{3.02{\text{ g}}}}{{1.00{\text{ g/mol}}}}\\&=3{\text{ mol}}\\\end{aligned}[/tex]
Since the reaction mixture has 1.0 mole of [tex]{{\text{N}}_2}[/tex] and 3.0 moles of [tex]{{\text{H}}_2}[/tex] and according to balance reaction 1 mole of [tex]{{\text{N}}_2}[/tex] and 3 moles of [tex]{{\text{H}}_2}[/tex] reacts to form 2 moles of [tex]{\text{N}}{{\text{H}}_3}[/tex].
The mass of 2 moles of ammonia [tex]\left({{\text{N}}{{\text{H}}_3}}\right)[/tex] can be calculated as,
[tex]\begin{aligned}{\text{Mass of N}}{{\text{H}}_3} &={\text{Moles of N}}{{\text{H}}_3} \times{\text{Molar mass of N}}{{\text{H}}_3}\\&=2.0{\text{ mol}}\times 17.03\,{\text{g/mol}}\\&={\mathbf{34}}{\mathbf{.06 g}}\\\end{aligned}[/tex]
Learn more:
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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, 0.280 mol of N2, 0.884 mol of H2, 0.044 mol H2.
