Question

under certain conditions, the substance ammonium nitrate can be broken down to form dinitrogen monoxide and water. if 22.8 grams of ammonium nitrate react to form 12.5 grams of dinitrogen monoxide, how many grams of water must simultaneously be formed? grams water

Explanation:

Step1: Write the chemical equation

The decomposition of ammonium nitrate is \(NH_4NO_3
ightarrow N_2O + 2H_2O\).

Step2: Calculate the molar - mass of relevant substances

The molar - mass of \(NH_4NO_3\) is \(M_{NH_4NO_3}=(14 + 4\times1+14 + 3\times16)=80\ g/mol\), the molar - mass of \(N_2O\) is \(M_{N_2O}=(2\times14 + 16)=44\ g/mol\), and the molar - mass of \(H_2O\) is \(M_{H_2O}=(2\times1+16)=18\ g/mol\).

Step3: Use the law of conservation of mass

According to the chemical equation, the mass ratio of \(NH_4NO_3\) to \(H_2O\) is \(80:(2\times18)=80:36\).
We know the mass of \(NH_4NO_3\) is \(m_{NH_4NO_3}=22.8\ g\). Let the mass of \(H_2O\) be \(m_{H_2O}\).
Set up the proportion \(\frac{m_{NH_4NO_3}}{m_{H_2O}}=\frac{80}{36}\).
Substitute \(m_{NH_4NO_3}=22.8\ g\) into the proportion: \(m_{H_2O}=\frac{22.8\times36}{80}\).

Step4: Calculate the mass of water

\(m_{H_2O}=\frac{22.8\times36}{80}=\frac{820.8}{80}=10.26\ g\).

Answer:

10.26 grams