- Determine the molar mass of oxygen: \(32.00 \text{ g/mol}\).
- Find the moles of oxygen: \(n(\text{O}_2) = \frac{4.53}{32.00} = 0.1416 \text{ mol}\).
- Calculate the moles of propane needed: \(n(\text{C}_3\text{H}_8) = \frac{0.1416}{5} = 0.0283 \text{ mol}\).
- Multiply by the molar mass of propane (\(44.094 \text{ g/mol}\)) to get the final mass: \(\boxed{1.25 \text{ g}}\).
Explanation
- Analyzing the Reaction and Given Data
The combustion reaction for propane (\(\text{C}_3\text{H}_8\)) is given by:
\[\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\]
From this balanced equation, we see that 1 mole of propane reacts with 5 moles of oxygen gas (\(\text{O}_2\)). We are given \(4.53 \text{ g}\) of oxygen and need to determine the mass of propane required for complete combustion. We use atomic masses \(C \approx 12.01 \text{ g/mol}\), \(H \approx 1.008 \text{ g/mol}\), and \(O \approx 16.00 \text{ g/mol}\).
- Calculating Moles of Oxygen
First, calculate the molar mass of oxygen gas (\(\text{O}_2\)):
\[M(\text{O}_2) = 2 \times 16.00 \text{ g/mol} = 32.00 \text{ g/mol}\]
Next, determine the number of moles of oxygen in \(4.53 \text{ g}\):
\[n(\text{O}_2) = \frac{4.53 \text{ g}}{32.00 \text{ g/mol}} = 0.1415625 \text{ mol}\]
- Calculating Moles and Molar Mass of Propane
Based on the stoichiometric coefficients of the balanced equation, 1 mole of \(\text{C}_3\text{H}_8\) requires 5 moles of \(\text{O}_2\). Therefore, the number of moles of propane required is:
\[n(\text{C}_3\text{H}_8) = \frac{n(\text{O}_2)}{5} = \frac{0.1415625}{5} = 0.0283125 \text{ mol}\]
Now, calculate the molar mass of propane (\(\text{C}_3\text{H}_8\)):
\[M(\text{C}_3\text{H}_8) = 3 \times 12.01 + 8 \times 1.008 = 36.03 + 8.064 = 44.094 \text{ g/mol}\]
- Determining the Final Mass
Finally, calculate the required mass of propane:
\[m(\text{C}_3\text{H}_8) = n(\text{C}_3\text{H}_8) \times M(\text{C}_3\text{H}_8) = 0.0283125 \text{ mol} \times 44.094 \text{ g/mol} \approx 1.2484 \text{ g}\]
Rounding this result to the closest value provided in the options gives \(1.25 \text{ g}\). This corresponds to option (d). Therefore, the mass of propane necessary to react with \(4.53 \text{ g}\) of oxygen is \(\boxed{1.25 \text{ g}}\).
Examples
Understanding stoichiometry is essential in real-life applications like chemical engineering, where engineers must calculate the exact amounts of reactants required for industrial processes. For example, in an automobile engine, knowing the precise fuel-to-air ratio ensures that the engine burns gasoline efficiently, maximizing power while minimizing waste and harmful emissions. The calculation we performed follows these exact principles of mass conservation and balanced chemical proportions.