Question

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

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

1. Write the balanced chemical equation

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

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 \(= \frac{500 L}{22.4 \, L/mol} ≈ 22.32 \, moles\).

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.

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: \[22.32 \, moles \, O_2 × \frac{4}{3} ≈ 29.76 \, moles \, H_2O\]

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.

Key concepts

Understanding Stoichiometry

Stoichiometry is like a recipe for a chemical reaction. It tells us how much of each ingredient (reactant) we need and how much product we can expect. For our reaction, the equation is:

• \(2 \text{CH}_3\text{OH(l)} + 3 \text{O}_2\text{(g)} \rightarrow 2 \text{CO}_2\text{(g)} + 4 \text{H}_2\text{O(g)}\)

This balanced equation indicates that two molecules of methanol react with three molecules of oxygen to produce two molecules of carbon dioxide and four molecules of water. Stoichiometry helps us predict the amounts of reactants and products involved in this reaction by comparing their ratios as seen in the equation.
To apply this to the problem, we follow the stoichiometric coefficients from the balanced chemical equation, which are crucial in calculating how much of each substance is needed or produced.

Exploring Limiting Reactant

In every reaction, the limiting reactant is the substance that gets completely used up first, stopping the reaction from continuing. It basically decides how much product can be formed. Let's consider our exercise:

• The balanced equation shows the mole ratio needed between methanol and oxygen.
• Methanol requires three moles of oxygen for every two moles of methanol.

Here, comparing the given moles of methanol and oxygen with the ratio from the equation helps identify the limiting reactant.
Calculations showed that oxygen has fewer moles than needed to react with all the methanol, making oxygen the limiting factor in this combustion reaction. Once the limiting reactant is used up, no further reaction can occur, determining the maximum amount of product formed.

Molar Volume in Chemistry

The concept of molar volume is pivotal when dealing with gases. At standard temperature and pressure (STP), one mole of any gas occupies a volume of 22.4 liters. This simplified relation makes calculations involving gas substances easier.

• For oxygen, we were given a volume at STP.
• Using the molar volume of 22.4 L/mol, we calculated moles of oxygen.

The result showed us that 500 liters of oxygen roughly equate to 22.32 moles.
Molar volume helps bridge the gap between measuring gas volume and understanding molecular quantities, critical for subsequent stoichiometric calculations.

Understanding Density Calculations

Density calculations are essential for converting between mass and volume of a substance, especially if it changes state within a reaction. In this exercise:

• Methanol's density allowed us to calculate its mass, given its volume in liters.
• This mass was then converted to moles for stoichiometric calculations.

Similarly, for water, knowing its density, we converted moles back to mass and finally to liquid volume:
- The density of water is approximately 1 g/mL,
- Using this, we calculated that 29.76 moles of water forms 536.1 mL of liquid.
Such conversions are useful, especially when the reaction product needs to be collected as a liquid, as they help estimate the final volume of the product.