Chemistry, 25.07.2019 07:30 ramentome7542

When a 3.80 g sample of magnesium nitride (mw 101g/mol) is reacted with 3.30 g of water, 3.60 g of mgo is formed. what is the percent yield of this reaction? mg3n2 + 3 h2o --> 2 nh3 + 3 mgo

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Answer from: dbuchlerdb5599

The percent yield of the reaction is $\boxed{{\text{79}}{\text{.19 \% }}}$ .

Further Explanation:

Balanced chemical reaction:

The chemical reaction that contains equal number of atoms of the different elements in the reactant as well as in the product side is known as balanced chemical reaction. The chemical equation is required to be balanced to follow the Law of the conservation of mass.

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,

${\text{A}} + 2{\text{B}} \to 3{\text{C}}$

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. 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.

Percent yield:

Percentage yield is defined as the percent ratio of actual yield to theoretical yield. The formula of the percentage yield is actual yield divided by theoretical yield and multiplied by 100.

${\text{Percent yield }}\left(\%\right)=\left( {\frac{{{\text{actual yield}}}}{{{\text{theoretical yield}}}}} \right) \times {\text{100\% }}$ ......(1)

Actual yield is the amount of actual product formed in a chemical reaction.

Theoretical yield is the amount of product formed from a reactant on the basis of stoichiometry in a balanced chemical reaction.

The balanced chemical reaction between ${{\text{H}}_{\text{2}}}{\text{O}}$ and is as follows:

${\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_2}+{\text{2}}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{2N}}{{\text{H}}_3} + 3{\text{MgO}}$ ……. (2)

The formula to determine number of moles is as follows:

${\text{number of moles }}=\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$

......(3)

Given mass of ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_2}$ is $3.8\;{\text{g}}$ .

Molar mass of ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_2}$ is $101\;{\text{g/mol}}$ .

Substitute these values in equation (3) to calculate number of moles of ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_2}$ .

$\begin{aligned}{\text{Number of moles}}&=\frac{{{\text{3}}{\text{.8 g}}}}{{{\text{101 g/mol}}}}\\&= {\text{0}}{\text{.03762 mol}}\\\end{aligned}$

The number of moles of ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_2}$ calculated is ${\text{0}}{\text{.03762 mol}}$ .

In accordance to the stoichiometry of reaction (2), 1 mole of ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_2}$ produce 3 moles of MgO.

Therefore the number of moles of MgO will be equal to thrice of the number of moles of ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_2}$ and calculated as follows:

$\begin{aligned}{\text{Moles of MgO}}&= 3\left( {{\text{0}}{\text{.03762 mol}}} \right)\\&= 0.1128{\text{ mol}}\\\end{aligned}$

The formula to calculate the mass of MgO is as follows:

${\text{Mass}} = {\text{number of moles}} \times {\text{molar mass}}$

......(5)

Molar mass of MgO is ${\text{40}}{\text{.304 g/mol}}$ .

Number of moles of MgO is $0.1128{\text{ mol}}$ .

Substitute these values in equation (4) to calculate mass of MgO.

$\begin{aligned}{\text{mass}}&= \left( {0.1128{\text{ mol}}} \right)\left( {\frac{{{\text{40}}{\text{.304 g}}}}{{{\text{1 mol}}}}} \right)\\&= 4.5462{\text{ g}}\\\end{aligned}$

The mass of MgO is $4.5462{\text{ g}}$ .

The actual mass of MgO is $3.60\;{\text{g}}$ .

The theoretical mass of MgO is $4.5462{\text{ g}}$ .

Substitute these values in equation (1) to calculate the percent yield.

$\begin{aligned}{\text{Percent yield}}\left( \% \right)&= \left( {\frac{{3.60\;{\text{g}}}}{{4.5462{\text{ g}}}}} \right){\text{100}}\;\%\\&= 79.1870{\text{ \% }}\\&\approx {\text{79}}{\text{.19 \% }}\\\end{aligned}$

The percent yield of the reaction is ${\mathbf{79}}{\mathbf{.19 \% }}$ .

Learn more:

1. Calculation of the percentage ionization of nitrous acid: https: .

2. Calculation of the percentage yield of reaction:

Answer details:

Grade: High school

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: Chemical reaction, reactant, product, number of moles, balanced chemical reaction, stoichiometry, percent yield, actual yield, theoretical yield and 71.19%.

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