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Chapter 9: Problem 48

How many liters \(\mathrm{N}_{2},\) density \(0.92 \mathrm{g} / \mathrm{L},\) can be made by the decomposition of 2.05 \(\mathrm{g} \mathrm{NaN}_{3} ?\) $$2 \mathrm{NaN}_{3}(s) \rightarrow 2 \mathrm{Na}(s)+3 \mathrm{N}_{2}(g)$$

Short Answer

Expert verified
The volume of nitrogen gas that can be made by the decomposition of 2.05 g sodium azide is approximately 1.44 liters.

Step by step solution

01

Calculate the moles of sodium azide

Calculate the number of moles of sodium azide using its molar mass. The molar mass of sodium azide (\( \mathrm{NaN}_{3} \)) is approximately 65 g/mol. Hence, the moles of sodium azide can be obtained as follows: \[ \mathrm{moles}=\frac{\mathrm{mass}}{\mathrm{Molar mass}}=\frac{2.05 g}{65 g/mol}=0.03154 mol \]
02

Calculate the moles of nitrogen gas

From stoichiometry, 2 moles of sodium azide produce 3 moles of nitrogen gas. Therefore the moles of nitrogen gas produced can be calculated as: \[ \mathrm{moles} \, \mathrm{of} \, \mathrm{N}_{2}=1.5 \times \mathrm{moles} \, \mathrm{of} \, \mathrm{NaN}_{3} =1.5 \times 0.03154\, \mathrm{mol} =0.04731 \, \mathrm{mol} \]
03

Calculate the volume of nitrogen gas

To calculate the volume of nitrogen using the density and the relationship: \[ \mathrm{Density} = \frac{\mathrm{mass}}{\mathrm{volume}} \] Therefore, rearranging for volume: \[ \mathrm{Volume} = \frac{\mathrm{mass}}{\mathrm{Density}} \] The mass of nitrogen can be calculated by multiplying the number of moles of nitrogen with molar mass. Since the molar mass of nitrogen is approximately 28g/mol, that gives: \[ \mathrm{mass} = \mathrm{moles} \times \mathrm{Molar \, mass} =0.04731 \, \mathrm{mol} \times 28 \, \mathrm{g/mol} =1.32468 \, \mathrm{g} \] Substituting the mass and the density into the volume formula gives: \[ \mathrm{Volume}=\frac{\mathrm{mass}}{\mathrm{Density}}=\frac{1.32468 g}{0.92 g/L}=1.44L \]

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Most popular questions from this chapter

How many grams of \(\mathrm{NaNO}_{2}\) form when 256 \(\mathrm{g} \mathrm{NaNO}_{3}\) react? The yield is 91\(\% .\) $$2 \mathrm{NaNO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{2}(s)+\mathrm{O}_{2}(g)$$

Why is a balanced chemical equation needed to solve stoichiometry problems?

How many grams \(\mathrm{CO}_{2}\) form from the complete combustion of \(1.00 \mathrm{L} \mathrm{C}_{8} \mathrm{H}_{18},\) density 0.700 \(\mathrm{g} / \mathrm{mL} ?\) If only \(1.90 \times 10^{3} \mathrm{g} \mathrm{CO}_{2}\) form, what is the percentage yield?

The chemical equation for the formation of water is $$2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}$$ \begin{equation}\begin{array}{l}{\text { a. If } 3.3 \mathrm{mol} \mathrm{O}_{2} \text { are used, how many moles }} \\ {\text { of } \mathrm{H}_{2} \text { are needed? }} \\ {\text { b. How many moles } \mathrm{O}_{2} \text { must react with }} \\ {\text { excess } \mathrm{H}_{2} \text { to form } 6.72 \mathrm{mol} \mathrm{H}_{2} \mathrm{O} \text { ? }} \\\ {\text { c. If you wanted to make } 8.12 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},} \\ {\text { how many moles of } \mathrm{H}_{2} \text { would you need? }}\end{array}\end{equation}

Write a balanced equation for the combus- tion of octane, \(\mathrm{C}_{8} \mathrm{H}_{18},\) with oxygen to obtain carbon dioxide and water. What is the mole ratio of oxygen to octane?
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