Question

Using the following unbalanced equation, answer the following questions?
$N{H_4}N{O_3}\left( s \right)\xrightarrow{\Delta }{N_2}O\left( g \right) + {H_2}O\left( g \right)$
(A) What is the mole ratio of $N{H_4}N{O_3}$ to ${H_2}O$ ?
(B) How many grams of water are produced if $33.0$ grams of ${N_2}O$ is produced?
(C) How many moles of ${N_2}O$ are produced if you have $25$ grams of $N{H_4}N{O_3}$ ?
(D) How many moles of ${H_2}O$ are produced if you have $5$ moles of $N{H_4}N{O_3}$ ?
(E) How many grams of ${N_2}O$ are produced if you have $125$ grams of $N{H_4}N{O_3}$ ?

Answer 1

From the balanced equation, the number of moles that can be produced from one mole of reactant. The number of moles is the ratio of the mass of the substance to the molar mass of the substance. By converting the moles and grams, the mass of other substances can be produced.

Complete answer:
Ammonium nitrate is a chemical compound formed by the combination of ammonium ion and nitrate ion. On decomposition, ammonium nitrate decomposes into nitrous oxide and water.
The balanced decomposition reaction of ammonium nitrate is:
$N{H_4}N{O_3}\left( s \right)\xrightarrow{\Delta }{N_2}O\left( g \right) + 2{H_2}O\left( g \right)$
From the above equation, it was clear that one mole of ammonium nitrate produces one mole of nitrous oxide and two moles of water.
Molar mass of $N{H_4}N{O_3}$ is $80gmo{l^{ - 1}}$
Molar mass of ${N_2}O$ is $44gmo{l^{ - 1}}$
Molar mass of ${H_2}O$ is $18gmo{l^{ - 1}}$
(A) Mole ratio of $N{H_4}N{O_3}$ to ${H_2}O$ is $1:2$
(B) Moles of ${N_2}O$ is produced will be $\dfrac{{33g}}{{44gmo{l^{ - 1}}}} = 0.75moles$
Moles of ${H_2}O$ produced will be $2 \times 0.75 = 1.5moles$
Grams of ${H_2}O$ produced will be $1.5 \times 18 = 27g$
Thus, $27g$ of water are produced from $33.0$ grams of ${N_2}O$
(C) Moles of $25$ grams of $N{H_4}N{O_3}$ will be $\dfrac{{25g}}{{80gmo{l^{ - 1}}}} = 0.3125moles$
As one mole of $N{H_4}N{O_3}$ produce one mole of ${N_2}O$
Thus, $0.3125moles$ of ${N_2}O$ are produced from $25$ grams of $N{H_4}N{O_3}$
(D) As one mole of $N{H_4}N{O_3}$ produce one mole of ${H_2}O$
Thus, $5$ moles of $N{H_4}N{O_3}$ produces $10$ moles of ${H_2}O$
(E) From part c, $0.3125moles$ of ${N_2}O$ are produced from $25$ grams of $N{H_4}N{O_3}$
Thus, $0.3125 \times 5 = 1.5625moles$ of ${N_2}O$ are produced from $125$ grams of $N{H_4}N{O_3}$
The grams of ${N_2}O$ produced will be $1.5625 \times 44 = 68.75g$
Thus, $68.75g$ of ${N_2}O$ are produced from $125$ grams of $N{H_4}N{O_3}$ .

Note:
The molar mass of the above chemical substances must be taken accurately. All the atoms present in chemical compounds molar mass should be taken based on the periodic table only. The number of moles is the ratio of mass to molar mass. Thus, conversions should be made properly from moles to grams.