Question

1. given 9.98 grams of n₂, how many moles of nan₃ are produced?

2 nan₃ → 2 na + 3 n₂

1. given 7.85 moles of nan₃, how many grams of na are produced?

Explanation:

Question 7

Step1: Calculate moles of \(N_2\)

The molar mass of \(N_2\) is \(M = 2\times14.01\space g/mol=28.02\space g/mol\).
Using the formula \(n=\frac{m}{M}\), where \(m = 9.98\space g\) and \(M = 28.02\space g/mol\), we have \(n_{N_2}=\frac{9.98\space g}{28.02\space g/mol}\approx0.356\space mol\)

Step2: Use mole - ratio from the balanced equation

From the balanced equation \(2NaN_3
ightarrow2Na + 3N_2\), the mole ratio of \(NaN_3\) to \(N_2\) is \(\frac{n_{NaN_3}}{n_{N_2}}=\frac{2}{3}\)
So \(n_{NaN_3}=\frac{2}{3}\times n_{N_2}\)
Substitute \(n_{N_2} = 0.356\space mol\) into the equation: \(n_{NaN_3}=\frac{2}{3}\times0.356\space mol\approx0.237\space mol\)

Step1: Use mole - ratio from the balanced equation

From the balanced equation \(2NaN_3
ightarrow2Na + 3N_2\), the mole ratio of \(Na\) to \(NaN_3\) is \(\frac{n_{Na}}{n_{NaN_3}}=\frac{2}{2} = 1\)
Given \(n_{NaN_3}=7.85\space mol\), then \(n_{Na}=7.85\space mol\)

Step2: Calculate mass of \(Na\)

The molar mass of \(Na\) is \(M = 22.99\space g/mol\)
Using the formula \(m=n\times M\), where \(n = 7.85\space mol\) and \(M = 22.99\space g/mol\)
\(m_{Na}=7.85\space mol\times22.99\space g/mol\approx180.5\space g\)

Answer:

\(0.237\space mol\)

Question 8