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Chapter 9: Problem 21

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}

Short Answer

Expert verified
a. 6.6 moles of \(H_2\) are needed for 3.3 moles of \(O_2\). b. 3.36 moles of \(O_2\) must react with excess \(H_2\) to form 6.72 moles of \(H_{2}O\). c. 16.24 moles of \(H_2\) would be needed to make 8.12 moles of \(H_{2}O\).

Step by step solution

01

Analyze the molar ratio of \(H_2\) and \(O_2\) in the chemical reaction

The chemical equation given is \(2 H_{2}+O_{2} \rightarrow 2 H_{2} O\). The coefficients in this equation tell us that for every 1 mole of \(O_2\) used, 2 moles of \(H_2\) and \(H_{2}O\) are produced. This is the molar ratio of \(H_2\), \(O_2\), and \(H_{2}O\).
02

Calculate the number of moles of \(H_2\) needed for 3.3 moles of \(O_2\)

The molar ratio obtained in step 1 indicates that for every 1 mole of \(O_2\), 2 moles of \(H_2\) are needed. Therefore, for 3.3 moles of \(O_2\), the required moles of \(H_2\) would be the molar ratio times the given moles of \(O_2\). Calculating, \[2 \times 3.3 = 6.6\] moles of \(H_2\) are needed.
03

Calculate the number of moles of \(O_2\) to react with excess \(H_2\) to form 6.72 moles of \(H_{2}O\)

The molar ratio also indicates that for every 1 mole of \(O_2\), 2 moles of \(H_{2}O\) are produced. Hence, if 6.72 moles of \(H_{2}O\) are to be formed, the required moles of \(O_2\) would be the given moles of \(H_{2}O\) divided by the molar ratio, which is \(6.72 \, \text{moles} / 2 = 3.36 \, \text{moles of} \, O_2\).
04

Calculate the number of moles of \(H_2\) needed to make 8.12 moles of \(H_{2}O\)

To make 8.12 moles of \(H_{2}O\), the required moles of \(H_2\) would be the given moles of \(H_{2}O\) times the molar ratio. So, \[2 \times 8.12 = 16.24 \] moles of \(H_2\) would be needed.

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Most popular questions from this chapter

The following reaction can be used to remove \(\mathrm{CO}_{2}\) breathed out by astronauts in a spacecraft. $$2 \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)$$ \begin{equation}\begin{array}{l}{\text { a. How many grams of carbon dioxide can }} \\ {\text { be removed by } 5.5 \text { mol LiOH? }} \\ {\text { b. How many milliliters } \mathrm{H}_{2} \mathrm{O} \text { (density }=} \\\ {0.997 \mathrm{g} / \mathrm{mL} \text { ) could form from } 25.7 \mathrm{g} \text { LiOH? }} \\ {\text { c. How many molecules } \mathrm{H}_{2} \mathrm{O} \text { could be }} \\ {\text { made when } 3.28 \mathrm{g} \mathrm{CO}_{2} \text { react? }}\end{array}\end{equation}

Oxygen can be prepared by heating potassium chlorate. $$2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$$ \begin{equation}\begin{array}{l}{\text { a. What mass of } \mathrm{O}_{2} \text { can be made from heat- }} \\ {\text { ing } 125 \text { g of } \mathrm{KClO}_{3} ?} \\ {\text { b. How many grams of } \mathrm{KClO}_{3} \text { are needed }} \\ {\text { to make } 293 \mathrm{g} \mathrm{g}_{2} \text { ? }} \\ {\text { c. How many grams of KCl could form from }} \\\ {20.8 \mathrm{g} \mathrm{KClO}_{3} ?}\end{array}\end{equation}

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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 \(\mathrm{Na}_{2} \mathrm{O}\) ? $$\mathrm{Na}_{2} \mathrm{O}(s)+2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow 2 \mathrm{NaHCO}_{3}(s)$$

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