Respuesta :
30.6 g
First, determine the molar masses of the reactants and products. Start by looking up the atomic weights of all involved elements.
Atomic weight nitrogen = 14.0067
Atomic weight hydrogen = 1.00794
Molar mass N2 = 2 * 14.0067 = 28.0134 g/mol
Molar mass H2 = 2 * 1.00794 = 2.01588 g/mol
Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol
Now determine how many moles of each reactant we have
Moles nitrogen gas = 25.2 g / 28.0134 g/mol = 0.899569492 mol
Moles hydrogen gas = 8.42 g / 2.01588 g/mol = 4.176835923 mol
Determine the limiting reactant. The balanced equation shows that for every mole of nitrogen gas, we need 3 moles of hydrogen gas. Since we have over 4 moles of hydrogen gas and less than 1 mole of nitrogen gas, it's obvious that the limiting reactant we have is nitrogen gas. And for every mole of nitrogen gas used, we produce 2 moles of ammonia, so the expected yield is
2 * 0.899569492 mol = 1.799138983 mol.
Now we just need to multiply the number of moles of ammonia by the molar mass of ammonia, so
1.799138983 mol * 17.03052 g/mol = 30.64027244 g
Rounding to 1 decimal place gives 30.6 g
[tex]\boxed{30.62{\text{ g}}}[/tex]of [tex]{\text{N}}{{\text{H}}_3}[/tex] will be produced from 25.2 g of [tex]{{\text{N}}_{\text{2}}}[/tex] and 8.42 g of [tex]{{\text{H}}_{\text{2}}}[/tex].
Further explanation:
Limiting reagent:
It is completely consumed in a chemical reaction. It decides the amount of product formed in any chemical 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 [tex]{\text{N}}{{\text{H}}_3}[/tex] is as follows:
[tex]3{{\text{H}}_2}\left(g\right)+{{\text{N}}_2}\left(g\right)\to2{\text{N}}{{\text{H}}_3}\left(g\right)[/tex]
From the balanced chemical reaction, the reaction stoichiometry between [tex]{{\text{H}}_{\text{2}}}[/tex] and[tex]{{\text{N}}_2}[/tex] is as follows:
[tex]\boxed{3{\text{ mol }}{{\text{H}}_2}:1{\text{ mol }}{{\text{N}}_2}}[/tex]
According to the reaction stoichiometry, 3 moles of [tex]{{\text{H}}_{\text{2}}}[/tex] and 1 mole of [tex]{{\text{N}}_2}[/tex] react to form 2 moles of [tex]{\text{N}}{{\text{H}}_3}[/tex]but the reaction mixture has 25.2 g of [tex]{{\text{N}}_2}[/tex] and 8.42 g of [tex]{{\text{H}}_{\text{2}}}[/tex].
The formula to calculate the moles of [tex]{{\text{H}}_{\text{2}}}[/tex] is as follows:
[tex]{\text{Moles of }}{{\text{H}}_{\text{2}}}=\frac{{{\text{Given mass of }}{{\text{H}}_{\text{2}}}}}{{{\text{Molar mass of }}{{\text{H}}_{\text{2}}}}}[/tex] ...... (1)
Substitute 8.42 g for the given mass of [tex]{{\text{H}}_{\text{2}}}[/tex] and 2.02 g/mol for the molar mass of [tex]{{\text{H}}_{\text{2}}}[/tex] in equation (1).
[tex]\begin{aligned}{\text{Moles of}}{{\text{H}}_{\text{2}}}&=\left({{\text{8}}{\text{.42 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{2}}{\text{.02 g}}}}}\right)\\&=4.168{\text{ mol}}\\\end{aligned}[/tex]
The formula to calculate the moles of [tex]{{\text{N}}_2}[/tex] is as follows:
[tex]{\text{Moles of}}{{\text{N}}_2}=\frac{{{\text{Given mass of}}{{\text{N}}_2}}}{{{\text{Molar mass of }}{{\text{N}}_2}}}[/tex] ...... (2)
Substitute 25.2 g for the given mass of [tex]{{\text{N}}_2}[/tex] and 28.01 g/mol for the molar mass of [tex]{{\text{N}}_2}[/tex] in equation (2).
[tex]\begin{aligned}{\text{Moles of }}{{\text{N}}_2}&=\left({{\text{25}}{\text{.2 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{28}}{\text{.01 g}}}}} \right)\\&=0.899{\text{ mol}}\\\end{aligned}[/tex]
Calculate the moles of [tex]{{\text{N}}_2}[/tex] that react with and 4.168 moles of [tex]{{\text{H}}_{\text{2}}}[/tex]as follows:
[tex]\begin{aligned}{\text{Amount of }}{{\text{N}}_2}\left( {{\text{mol}}}\right)&=\left({{\text{4}}{\text{.168 mol }}{{\text{H}}_2}}\right)\left({\frac{{1{\text{mol}}{{\text{N}}_2}}}{{{\text{3 mol }}{{\text{H}}_2}}}}\right)\\&=1.389{\text{mol}}{{\text{N}}_2}\\\end{aligned}[/tex]
The calculations concluded that 1.389 moles of [tex]{{\text{N}}_2}[/tex]is required for 4.168 moles of [tex]{{\text{H}}_{\text{2}}}[/tex] but we have only 0.899 moles of [tex]{{\text{N}}_2}[/tex] in the reaction mixture. Therefore, [tex]{{\mathbf{N}}_{\mathbf{2}}}[/tex]is present in limited quantity and is a limiting reagent.
According to the balanced chemical equation 1 mole of [tex]{{\text{N}}_2}[/tex]produce 2 moles of [tex]{\text{N}}{{\text{H}}_3}[/tex], therefore, the moles of [tex]{\text{N}}{{\text{H}}_3}[/tex]produced by 1.389 moles of [tex]{{\text{N}}_2}[/tex] can be calculated as follows:
[tex]\begin{aligned}{\text{Amount of N}}{{\text{H}}_3}\left({{\text{mol}}}\right)&=\left({{\text{0}}{\text{.899 mol}}{{\text{N}}_2}}\right)\left({\frac{{2{\text{mol N}}{{\text{H}}_3}}}{{1{\text{ mol }}{{\text{N}}_2}}}}\right)\\&=1.798{\text{mol N}}{{\text{H}}_{\text{3}}}\\\end{aligned}[/tex]
The formula to calculate the mass of [tex]{\text{N}}{{\text{H}}_3}[/tex] is as follows:
[tex]{\text{Mass of N}}{{\text{H}}_3}=\left({{\text{Moles of N}}{{\text{H}}_3}}\right)\left({{\text{Molar mass of N}}{{\text{H}}_3}}\right)[/tex] ...... (3)
Substitute 1.798 mol for the moles of [tex]{\text{N}}{{\text{H}}_3}[/tex] and 17.03 g/mol for the molar mass of [tex]{\text{N}}{{\text{H}}_3}[/tex] in equation (3).
[tex]\begin{aligned}{\text{Mass of N}}{{\text{H}}_3}&=\left({{\text{1}}{\text{.798 mol}}}\right)\left({\frac{{{\text{17}}{\text{.03 g}}}}{{{\text{1 mol}}}}}\right)\\&=30.619{\text{ g}}\\&\approx 30.62{\text{ g}}\\\end{aligned}[/tex]
So 30.62 g of [tex]{\mathbf{N}}{{\mathbf{H}}_{\mathbf{3}}}[/tex] will be produced from 25.2 g of [tex]{{\mathbf{N}}_{\mathbf{2}}}[/tex] and 8.42 g of [tex]{{\mathbf{H}}_{\mathbf{2}}}[/tex].
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Mole concept
Keywords: limiting reagent,NH3,N2,H2,2NH3,3H2,30.62 g,2.778 mol,17.03 g/mol,2.02 g/mol,1.389 mol,25.2 g,8.42 g,4.168 mol,0.899 mol,1 mole,2 moles,3 moles.
