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Chapter 9: Problem 40

Assume that 44.3 \(\mathrm{g} \mathrm{Na}_{2} \mathrm{O}\) are formed during the inflation of an air bag. How many liters of \(\mathrm{CO}_{2}(\) density \(=1.35 \mathrm{g} / \mathrm{L})\) are needed to completely react with the \(\mathrm{Na}_{2} \mathrm{O}\) ? $$\mathrm{Na}_{2} \mathrm{O}(s)+2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow 2 \mathrm{NaHCO}_{3}(s)$$

Short Answer

Expert verified
The volume of CO2 needed to completely react with the Na2O is approximately 47.02 liters.

Step by step solution

01

Convert given mass of Na2O to moles

Use the molar mass of Na2O to convert grams to moles. The molar mass of Na2O is 61.98 g/mol. So, \(Moles of Na2O = \frac{44.3 g}{61.98 g/mol} = 0.714 mol\)
02

Find required moles of CO2 using stoichiometry

Stoichiometry of the balanced chemical reaction shows that 2 moles of CO2 react with one mole of Na2O. Therefore, the required moles of CO2 = 2 * moles of Na2O = 2 * 0.714 mol = 1.428 mol
03

Convert moles of CO2 to volume

Use the density of CO2 to do this. The density is given as 1.35 g/L which is the same as its mass divided by volume. Therefore rearranging, volume = mass/density. But we know mass = moles * Molar mass. Substituting this into our volume equation gives, volume = moles * Molar mass / density. Therefore, volume of CO2 = \( \frac{1.428 mol * 44.01 g/mol}{1.35 g/L} = 47.02 L \)

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Most popular questions from this chapter

How many liters of \(\mathrm{O}_{2},\) density \(1.43 \mathrm{g} / \mathrm{L},\) are needed for the complete combustion of \(1.00 \mathrm{L} \mathrm{C}_{8} \mathrm{H}_{18},\) density 0.700 \(\mathrm{g} / \mathrm{mL} ?\)

Use the equation provided to answer the questions that follow. The density of oxygen gas is 1.428 \(\mathrm{g} / \mathrm{L}\) . $$2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$$ \begin{equation}\begin{array}{l}{\text { a. What volume of oxygen can be made }} \\ {\text { from } 5.00 \times 10^{-2} \text { mol of } \mathrm{KClO}_{3} ?} \\ {\text { b. How many grams } \mathrm{KClO}_{3} \text { must react to }} \\\ {\text { form } 42.0 \mathrm{mL} \mathrm{O}_{2} \text { ? }} \\ {\text { c. How many milliliters of } \mathrm{O}_{2} \text { will form at }} \\ {\text { STP from } 55.2 \mathrm{g} \mathrm{KClO}_{3} ?}\end{array}\end{equation}

What is the function of the catalytic converter in the exhaust system?

Calcium carbide, \(\mathrm{CaC}_{2},\) reacts with water to form acetylene. $$\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s)$$ \begin{equation}\begin{array}{l}{\text { a. How many grams of water are needed to }} \\ {\text { react with } 485 \text { g of calcium carbide? }} \\\ {\text { b. How many grams of } \mathrm{CaC}_{2} \text { could make }} \\\ {\text { 23.6 g } \mathrm{C}_{2} \mathrm{H}_{2} \text { ? }} \\ {\text { c. If } 55.3 \mathrm{g} \mathrm{Ca}(\mathrm{OH})_{2} \text { are formed, how many }} \\ {\text { grams of water reacted? }}\end{array}\end{equation}

Describe what might happen if too much or too little gas generant is used in an air bag.
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