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Chapter 10: Problem 119

Consider the combustion reaction between \(1.00 \mathrm{~L}\) of liquid methanol (density \(=0.850 \mathrm{~g} / \mathrm{mL}\) ) and \(500 \mathrm{~L}\) of oxygen gas measured at STP. The products of the reaction are \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) Calculate the volume of liquid \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion and you condense the water vapor.

Short Answer

Expert verified
Approximately \(536.1 \, mL\) of liquid water is formed if the reaction goes to completion and the water vapor is condensed.

Step by step solution

01

1. Write the balanced chemical equation

The combustion reaction of methanol (\(CH_3OH\)) and oxygen (\(O_2\)) is given by: \[ 2CH_3OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 4H_2O(g) \]
02

2. Calculate the moles of methanol and oxygen

First, we need to convert the given amounts of methanol and oxygen into moles. For methanol, we use its volume and density to obtain the mass and then divide by the molar mass. Mass of methanol \(= 1.00 L × 0.850 \frac{g}{mL} × 1000 \frac{mL}{L} = 850 g\) The molar mass of methanol, \(CH_3OH = 12.01 + 15.999 + (4 × 1.008) = 32.04186 \, g/mol\) Moles of methanol \(= \frac{850 g}{32.04186 \, g/mol} = 26.54 \, moles\). For oxygen, since it is measured at STP, we can use the molar volume of an ideal gas, which is \(22.4 \, L/mol\): Moles of oxygen \(= \frac{500 L}{22.4 \, L/mol} ≈ 22.32 \, moles\).
03

3. Determine limiting reactant

To find which substance is the limiting reactant, compare the mole ratios with the stoichiometry of the balanced equation. The mole ratio for the methanol to oxygen is: \[\frac{26.54}{2}:\frac{22.32}{3} ≈ 13.27:7.44\] This means that there are more moles of methanol available than necessary for complete combustion of the given amount of oxygen. Thus, oxygen is the limiting reactant.
04

4. Calculate moles of water formed

With the limiting reactant identified, we can now calculate the moles of water formed using stoichiometry. From the balanced equation, \(3 \, moles \, O_2\) react completely with \(2 \, moles \, CH_3OH\) to produce \(4 \, moles \, H_2O\). This means that: \[1 \, mole \, O_2 \rightarrow \frac{4}{3} \, moles \, H_2O\] Therefore, the moles of water formed are: \[22.32 \, moles \, O_2 × \frac{4}{3} ≈ 29.76 \, moles \, H_2O\]
05

5. Convert moles of water to volume of liquid water

Finally, we can convert the moles of water to the volume of liquid water using the density of water (which is approximately \(1 \frac{g}{mL}\)). The molar mass of water is \(18.015 \, g/mol\). Mass of water: \(= 29.76 \, moles \, H_2O × 18.015 \frac{g}{mole} ≈ 536.1 \, g\) Using the density of water, we can now find the volume of liquid water formed: Volume of water \(= \frac{536.1 \, g}{1 \frac{g}{mL}} = 536.1 \, mL\). So, approximately \(536.1 \, mL\) of liquid water is formed if the reaction goes to completion and the water vapor is condensed.

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