Question

the chemical equation below shows the decomposition of ammonium nitrate $(nh_{4}no_{3})$.$nh_{4}no_{3}\to n_{2}o + 2h_{2}o$a chemist who is performing this reaction starts with 160.1 g of $nh_{4}no_{3}$. the molar mass of $nh_{4}no_{3}$ is 80.03 g/mol, the molar mass of water $(h_{2}o)$ is 18.01 g/mol. what mass, in grams, of $h_{2}o$ is produced?9.01 grams18.01 grams36.03 grams72.06 grams

Explanation:

Step1: Calculate moles of \(NH_4NO_3\)

Use the formula \(n=\frac{m}{M}\), where \(n\) is moles, \(m\) is mass, and \(M\) is molar mass.
\(n_{NH_4NO_3}=\frac{160.1\ g}{80.05\ g/mol}\approx2\ mol\)

Step2: Use mole ratio from chemical equation

From \(NH_4NO_3
ightarrow N_2O + 2H_2O\), mole ratio \(NH_4NO_3:H_2O = 1:2\). So \(n_{H_2O}=2\times n_{NH_4NO_3}\)
\(n_{H_2O}=2\times2\ mol = 4\ mol\)

Step3: Calculate mass of \(H_2O\)

Use \(m = n\times M\), \(M_{H_2O}=18.01\ g/mol\)
\(m_{H_2O}=4\ mol\times18.01\ g/mol = 72.04\ g\approx72.06\ g\)

Answer:

72.06 grams