Question

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

Answer 1

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: brainly.com/question/3128225.

2. Calculation of the percentage yield of reaction: brainly.com/question/6222572

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%.

Answer 2

  The  percent   yield  of  this   reaction  is  calculated  as  follows
  Mg3N2  +  3H2O  =2NH3  +  3Mgo

  calculate  the   theoretical  yield,
moles=mass/molar  mass
moles Mg3N2=  3.82  g/100g/mol=  0.0382  moles(limiting  regent)

moles of  H2o=  7.73g/18g/mol  = 0.429  moles ( in  excess_)

by  use   of  mole  ratio   between  Mg3N2  to  MgO  which  is  1:3   the    moles  of MgO =  0.0382  x3  = 0.1146  moles
mass =moles  x molar  mass

the  theoretical  mass  is therefore =  0.1146mole  x  40  g/mol  =  4.58 grams

The %  yield  =  actual  mass/theoretical  mass  x1000

=  3.60/4.584  x100=  78.5%