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Chapter 3: Problem 66

Certain race cars use methanol (CH \(_{3} \mathrm{OH}\), also called wood alcohol) as a fuel. The combustion of methanol occurs according to the following equation:$$2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$$In a particular reaction, 9.8 moles of \(\mathrm{CH}_{3} \mathrm{OH}\) are reacted with an excess of \(\mathrm{O}_{2}\). Calculate the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) formed.

Short Answer

Expert verified
The number of moles of \(\mathrm{H}_{2} \mathrm{O}\) formed is 19.6

Step by step solution

01

Identify the relevant stoichiometric ratio

From the balanced chemical equation, you can see that the ratio between the moles of methanol (\(\mathrm{CH}_{3} \mathrm{OH}\)) and water (\(\mathrm{H}_{2} \mathrm{O}\)) is 2 : 4, or simplifying, 1 : 2. It means for every 1 mole of methanol reacted, 2 moles of water are formed.
02

Use the identified ratio to compute the moles of water

We know that there are 9.8 moles of methanol reacted. We use our ratio to calculate the quantity of water produced. By multiplying the 9.8 moles of methanol by the stoichiometric ratio of 2 moles of water per mole of methanol, we get the number of moles of water produced.
03

Calculation

The calculation becomes \(9.8 \: \text{moles CH}_{3} \text{OH} \times \frac{2 \: \text{moles H}_{2} \text{O}}{1 \: \text{mole CH}_{3} \text{OH}} = 19.6 \: \text{moles H}_{2} \text{O}\).

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Most popular questions from this chapter

The following reaction is stoichiometric as written$$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}+\mathrm{NaOC}_{2}\mathrm{H}_{5} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{8}+\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{OH}+\mathrm{NaCl}$$ but it is often carried out with an excess of \(\mathrm{NaOC}_{2} \mathrm{H}_{5}\) to react with any water present in the reaction mixture that might reduce the yield. If the reaction shown was carried out with \(6.83 \mathrm{~g}\) of \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}\), how many grams of \(\mathrm{NaOC}_{2} \mathrm{H}_{5}\) would be needed to have a 50 percent molar excess of that reactant?

Which of the following has the greater mass: \(0.72 \mathrm{~g}\) of \(\mathrm{O}_{2}\) or 0.0011 mole of chlorophyll \(\left(\mathrm{C}_{55} \mathrm{H}_{72} \mathrm{MgN}_{4} \mathrm{O}_{5}\right) ?\)

One of the reactions that occurs in a blast furnace, where iron ore is converted to cast iron, is$$\mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \longrightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_{2}$$Suppose that \(1.64 \times 10^{3} \mathrm{~kg}\) of \(\mathrm{Fe}\) are obtained from a \(2.62 \times 10^{3} \mathrm{~kg}\) sample of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). Assuming that the reaction goes to completion, what is the percent purity of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the original sample?

For many years the recovery of gold-that is, the separation of gold from other materials-involved the use of potassium cyanide: $$4 \mathrm{Au}+8 \mathrm{KCN}+\mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \underset{4 \mathrm{KAu}(\mathrm{CN})_{2}+4 \mathrm{KOH}}{\longrightarrow}$$ What is the minimum amount of KCN in moles needed to extract \(29.0 \mathrm{~g}\) (about an ounce) of gold?

What mole ratio of molecular chlorine \(\left(\mathrm{Cl}_{2}\right)\) to molecular oxygen \(\left(\mathrm{O}_{2}\right)\) would result from the breakup of the compound \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) into its constituent elements?
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