Question

part i: review of calculations based on chemical reactions
for part i we will review how to write chemical reactions and how to use mole to mole ratios for
calculations.

1. ammonium nitrate is a common fertilizer and has been the cause of multiple disasters due to

improper handling. when heat is applied to ammonium nitrate, it decomposes into nitrogen gas,
oxygen gas, and water vapor.
a. write the balanced chemical reaction for the decomposition of solid ammonium nitrate. be
sure to include the states of the reactant and products.
b. based on your balanced reaction, what is the conversion factor that allows you to relate the
moles of ammonium nitrate to total moles of gas produced?
c. use this conversion factor to find how many moles of gas are produced from the
decomposition of 95 grams of ammonium nitrate.
d. what is the total number of gas molecules produced from the 95 grams of ammonium
nitrate?
e. calculate the amount of nitrogen gas produced from the 95 grams of ammonium nitrate.
f. calculate the amount of oxygen gas produced from the 95 grams of ammonium nitrate.

Explanation:

Step1: Write the balanced chemical equation

The chemical formula for ammonium nitrate is \(NH_4NO_3\). The decomposition reaction is:
$$2NH_4NO_3(s)\xrightarrow{\Delta}2N_2(g)+O_2(g)+4H_2O(g)$$

Step2: Determine the conversion factor

From the balanced equation, for every \(2\) moles of \(NH_4NO_3\), \((2 + 1+4)=7\) moles of gas (\(N_2\), \(O_2\), \(H_2O\)) are produced. So the conversion factor is \(\frac{7\space mol\space gas}{2\space mol\space NH_4NO_3}\)

Step3: Calculate moles of \(NH_4NO_3\)

The molar mass of \(NH_4NO_3\) is \(M=(14\times2)+(1\times4)+(16\times3)=80\space g/mol\). Using \(n=\frac{m}{M}\), for \(m = 95\space g\), \(n_{NH_4NO_3}=\frac{95\space g}{80\space g/mol}=1.1875\space mol\)

Step4: Calculate moles of gas

Using the conversion factor \(\frac{7\space mol\space gas}{2\space mol\space NH_4NO_3}\), \(n_{gas}=1.1875\space mol\space NH_4NO_3\times\frac{7\space mol\space gas}{2\space mol\space NH_4NO_3}=4.15625\space mol\)

Step5: Calculate number of gas molecules

Using \(N = n\times N_A\) (where \(N_A = 6.022\times10^{23}\space mol^{-1}\)), \(N=4.15625\space mol\times6.022\times 10^{23}\space mol^{-1}=2.504\times10^{24}\)

Step6: Calculate moles of \(N_2\)

From the balanced equation, mole ratio of \(NH_4NO_3\) to \(N_2\) is \(2:2 = 1:1\). So \(n_{N_2}=1.1875\space mol\)

Step7: Calculate moles of \(O_2\)

From the balanced equation, mole ratio of \(NH_4NO_3\) to \(O_2\) is \(2:1\). So \(n_{O_2}=1.1875\space mol\times\frac{1}{2}=0.59375\space mol\)

Answer:

a. \(2NH_4NO_3(s)\xrightarrow{\Delta}2N_2(g)+O_2(g)+4H_2O(g)\)
b. \(\frac{7\space mol\space gas}{2\space mol\space NH_4NO_3}\)
c. \(4.16\space mol\) (rounded to three significant figures)
d. \(2.50\times 10^{24}\)
e. \(1.19\space mol\) (rounded to three significant figures)
f. \(0.594\space mol\) (rounded to three significant figures)