Chapter 3, Problem 3.161QP

Interpretation Introduction

Interpretation:

The amount of magnesium nitride and magnesium oxide formed from the given condition has to be calculated.

Concept Introduction:

• Balanced chemical equation of a reaction is written according to law of conservation of mass.
• Equation for Number of moles of a substance, from its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

• Equation for Number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

• Mole ratio between the reactant and a product of a reaction are depends upon the coefficients of reactant and product in a balanced chemical equation.

Answer to Problem 3.161QP

The mass of magnesium nitride is $\text{8}\text{.335g}$.

The mass of magnesium oxide is $\text{25}\text{.66g}$.

Explanation of Solution

To determine: the balanced chemical equations for the given chemical reaction of formation of ${\text{NH}}_3$.

${\text{Mg}}_3{\text{N}}_2\text{+}6{\text{H}}_2\text{O}\to 3\text{Mg(OH}{)}_2+2{\text{NH}}_3$

In given reaction, Magnesium nitride reacts with Water to form magnesium hydroxide and ammonia.

Therefore,

The chemical equation for this reaction is,

${\text{Mg}}_3{\text{N}}_2\text{+}{\text{H}}_2\text{O}\to \text{Mg(OH}{)}_2+{\text{NH}}_3$

Balanced chemical equation of a reaction is written according to law of conservation of mass.

Therefore,

The total number of each atoms in the reactant side should equal to the total number of each atoms in the product side.

So, in order to balance a chemical equation, the coefficients of compounds or atoms are needed to be changed in such a way that total number of each atoms in the reactant side and the total number of each atoms in the product side is to become equal.

Hence,

The balanced equations for the given reactions are,

${\text{Mg}}_3{\text{N}}_2\text{+}6{\text{H}}_2\text{O}\to 3\text{Mg(OH}{)}_2+2{\text{NH}}_3$

To determine: the mass of ${\text{Mg}}_3{\text{N}}_2$ produced from the burned sample of $\text{21}\text{.496g}$ magnesium sample.

The mass of ${\text{Mg}}_3{\text{N}}_2$ produced is $\text{8}\text{.335g}$.

The mass of ${\text{Mg}}_3{\text{N}}_2$ produced from the $\text{21}\text{.496g}$ of magnesium sample in the burning reaction is equals to the mass of ${\text{Mg}}_3{\text{N}}_2$ used to produce 2.813g of ${\text{NH}}_3$ in the given reaction.

The mass of ammonia produced is given as 2.813g.

Equation for Number of moles of a substance, from its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

Therefore,

The number of moles of ammonia produced in the reaction is,

$\text{Number of moles}\text{=}\frac{\text{2}\text{.813g}}{\text{17}\text{.034g}}\text{=}\text{0}\text{.16514mol}$.

The balanced chemical equation of the reaction is,

${\text{Mg}}_3{\text{N}}_2\text{+}6{\text{H}}_2\text{O}\to 3\text{Mg(OH}{)}_2+2{\text{NH}}_3$

The mole ratio between ${\text{NH}}_3$ and ${\text{Mg}}_3{\text{N}}_2$ in the reaction is 2:1.

So, the ${\text{Mg}}_3{\text{N}}_2$ used to produce 2.813g of ${\text{NH}}_3$ is,

$\frac{\text{1}}{\text{2}}\text{×}\text{0}\text{.16514}\text{mol}\text{=}\text{0}\text{.08257}\text{mol}$.

Equation for Number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

Then,

The mass of ${\text{Mg}}_3{\text{N}}_2$ used to produce 2.813g of ${\text{NH}}_3$ is,

$\text{0}\text{.08257}\text{×}\text{100}\text{.95g}\text{=}\text{8}\text{.335g}$.

Hence,

The mass of ${\text{Mg}}_3{\text{N}}_2$ produced from the burned sample of $\text{21}\text{.496g}$ magnesium sample

is $\text{8}\text{.335g}$.

To determine: the mass of $\text{Mg}$presented in $\text{MgO}$product:

The mass of $\text{Mg}$ presented in $\text{MgO}$ is $\text{15}\text{.474g}$.

The number of moles of ${\text{Mg}}_3{\text{N}}_2$ produced is found that $\text{0}\text{.08257}\text{mol}$.

There are three Mg elements in ${\text{Mg}}_3{\text{N}}_2$, so the mole ratio between $\text{Mg}$ and ${\text{Mg}}_3{\text{N}}_2$ is 3:1.

So, the number of moles of $\text{Mg}$ is,

$\text{3}\text{×}\text{0}\text{.08257}\text{mol}\text{=}\text{0}\text{.2477}\text{mol}$

Equation for Number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

Then,

The mass of $\text{Mg}$ presented in ${\text{Mg}}_3{\text{N}}_2$ is,

$\text{0}\text{.2477}\text{×}\text{24}\text{.31g}\text{=}\text{6}\text{.022g}$.

The total mass of Mg presented in $\text{MgO}$ and ${\text{Mg}}_3{\text{N}}_2$ is given as 21.496g.

So,

The mass of $\text{Mg}$ presented in $\text{MgO}$ is,

$\text{24}\text{.496g}\text{-}\text{6}\text{.022g}\text{=}\text{15}\text{.474g}$.

To determine: the mass of $\text{MgO}$ produced from the burned sample of$\text{21}\text{.496g}$ magnesium sample:

The mass of $\text{MgO}$ produced is $\text{8}\text{.335g}$.

The mass of $\text{Mg}$ presented in $\text{MgO}$ is found that $\text{15}\text{.474g}$.

Equation for Number of moles of a substance, from its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

So, the number of moles of $\text{Mg}$ presented in $\text{MgO}$ is,

$\text{Number of moles}\text{=}\frac{\text{15}\text{.474g}}{\text{24}\text{.31g}}\text{=}\text{0}\text{.6365}\text{mol}$

There is only one Mg elements in $\text{MgO}$, so the mole ratio between $\text{Mg}$ and $\text{MgO}$ is 1:1.

So, the number of moles of $\text{MgO}$ is,

$\text{1}\text{×}\text{0}\text{.6365}\text{mol}\text{=}\text{0}\text{.6365}\text{mol}$

Equation for Number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

Then,

The mass of $\text{MgO}$ is,

$\text{0}\text{.6365}\text{×}\text{40}\text{.31g}\text{=}\text{25}\text{.66g}$.