Question

the molar mass of nitrogen gas (n₂) is (round to the nearest tenths place) g/mol part b how many grams of n₂ are produced if 90 g of sodium azide is used? a 58 g n₂ b 39 g n₂ c 26 g n₂ d 90 g n₂ part c what is the proportion of moles of sodium azide that are used compared to the moles of sodium produced? a for every 3 moles of sodiu zide used, 3 moles of sodium are

Explanation:

Step1: Write the balanced chemical equation

The decomposition of sodium azide (\(NaN_3\)) is \(2NaN_3
ightarrow 2Na + 3N_2\)

Step2: Calculate the molar mass of \(NaN_3\)

The molar mass of \(Na\) is \(22.99\ g/mol\) and of \(N\) is \(14.01\ g/mol\). So molar mass of \(NaN_3=22.99+(3\times14.01)=22.99 + 42.03=65.02\ g/mol\)

Step3: Calculate moles of \(NaN_3\)

Moles of \(NaN_3=\frac{mass}{molar\ mass}=\frac{90\ g}{65.02\ g/mol}\approx1.38\ mol\)

Step4: Use mole ratio from the balanced equation

From \(2NaN_3
ightarrow 2Na + 3N_2\), the mole ratio of \(NaN_3\) to \(N_2\) is \(2:3\). If \(n(NaN_3) = 1.38\ mol\), then \(n(N_2)=\frac{3}{2}\times n(NaN_3)=\frac{3}{2}\times1.38 = 2.07\ mol\)

Step5: Calculate mass of \(N_2\)

Molar mass of \(N_2=(2\times14.01)=28.02\ g/mol\). Mass of \(N_2=n\times molar\ mass=2.07\ mol\times28.02\ g/mol\approx58\ g\)

Answer:

A. \(58\ g\ N_2\)