Question

what mass in grams of nan₃ is required to produce 50.2 l of n₂ gas (density = 1.25 g/l ) according to the balanced chemical reaction: 2 nan₃(s) → 2 na(s) + 3 n₂(g)

Explanation:

Step1: Calculate the mass of \(N_2\)

Use the formula \(m=
ho V\) (where \(m\) is mass, \(
ho\) is density, \(V\) is volume).
Given \(
ho = 1.25\space g/L\) and \(V = 50.2\space L\), then \(m_{N_2}=
ho V=1.25\space g/L\times50.2\space L = 62.75\space g\)

Step2: Calculate the moles of \(N_2\)

Use the formula \(n=\frac{m}{M}\) (where \(n\) is moles, \(m\) is mass, \(M\) is molar mass). The molar mass of \(N_2\) is \(M = 28.02\space g/mol\).
\(n_{N_2}=\frac{m_{N_2}}{M_{N_2}}=\frac{62.75\space g}{28.02\space g/mol}\approx 2.24\space mol\)

Step3: Calculate the moles of \(NaN_3\)

From the balanced equation \(2NaN_3(s)\to2Na(s)+3N_2(g)\), the mole ratio of \(NaN_3\) to \(N_2\) is \(\frac{n_{NaN_3}}{n_{N_2}}=\frac{2}{3}\).
So \(n_{NaN_3}=\frac{2}{3}n_{N_2}\). Substituting \(n_{N_2} = 2.24\space mol\), we get \(n_{NaN_3}=\frac{2}{3}\times2.24\space mol\approx1.49\space mol\)

Step4: Calculate the mass of \(NaN_3\)

The molar mass of \(NaN_3\) is \(M_{NaN_3}=22.99 + 3\times14.01=65.02\space g/mol\)
Use the formula \(m = nM\), so \(m_{NaN_3}=n_{NaN_3}M_{NaN_3}=1.49\space mol\times65.02\space g/mol\approx96.9\space g\)

Answer:

\(96.9\space g\)