Butane, C4H10, reacts with oxygen, O2, to form water, H2O, and carbon dioxide, CO2, as shown in the following chemical equation:
2C4H10(g)+1302(g)→10H2O(g) +8CO2(g)
The coefficients in this equation represent mole ratios. Notice that the coefficient for water (10) is five times that of butane (2). Thus, the number of moles of water produced is five times the number of moles of butane that react.
Also, notice that the coefficient for butane (2) is one-fourth the coefficient of carbon dioxide (8). Thus, the number of moles of butane that react is one-fourth the number of moles of carbon dioxide that you produce.
But be careful! If you are given the mass of a compound, you must first convert to moles before applying these ratios.
Butane,
Review | Constants | Periodic Table
Now that you know the molar masses of the relevant compounds, you are ready to start solving stoichiometry problems. In general, the typical strategy is
- Convert from grams of compound X to moles of compound X using the molar mass of compound X.
- Convert from moles of compound X to moles of compound Y using the coefficients in the balanced chemical equation.
- Convert from moles of compound Y to grams of compound Y using the molar mass of compound Y.
▼ Part B
Calculate the mass of water produced when 6.65 g of butane reacts with excess oxygen.
Express your answer to three significant figures and include the appropriate units.
The Correct Answer and Explanation is :
To calculate the mass of water produced when 6.65 g of butane (C4H10) reacts with excess oxygen, follow these steps:
Step 1: Convert grams of butane (C4H10) to moles of butane
The molar mass of butane (C4H10) is calculated as follows:
[
\text{Molar mass of C4H10} = (4 \times 12.01 \, \text{g/mol}) + (10 \times 1.008 \, \text{g/mol}) = 58.12 \, \text{g/mol}
]
Now, use the given mass of butane (6.65 g) to find the number of moles:
[
\text{Moles of C4H10} = \frac{\text{Mass of C4H10}}{\text{Molar mass of C4H10}} = \frac{6.65 \, \text{g}}{58.12 \, \text{g/mol}} = 0.114 \, \text{mol}
]
Step 2: Use the mole ratio from the balanced equation
From the balanced chemical equation:
[
2C4H10(g) + 13O2(g) \rightarrow 10H2O(g) + 8CO2(g)
]
The mole ratio between butane (C4H10) and water (H2O) is 2:10, which simplifies to 1:5. This means for every 1 mole of butane, 5 moles of water are produced. Therefore, we can use this ratio to find the moles of water produced:
[
\text{Moles of H2O} = \left(0.114 \, \text{mol C4H10}\right) \times \left(\frac{5 \, \text{mol H2O}}{1 \, \text{mol C4H10}}\right) = 0.570 \, \text{mol H2O}
]
Step 3: Convert moles of water to grams of water
The molar mass of water (H2O) is:
[
\text{Molar mass of H2O} = (2 \times 1.008 \, \text{g/mol}) + (1 \times 16.00 \, \text{g/mol}) = 18.016 \, \text{g/mol}
]
Now, convert moles of water to grams:
[
\text{Mass of H2O} = \text{Moles of H2O} \times \text{Molar mass of H2O} = 0.570 \, \text{mol} \times 18.016 \, \text{g/mol} = 10.26 \, \text{g}
]
Final Answer:
The mass of water produced when 6.65 g of butane reacts with excess oxygen is 10.3 g (rounded to three significant figures).
Explanation:
This problem follows the general stoichiometry process. First, we converted the given mass of butane to moles. Then, we used the mole ratio from the balanced equation to find the moles of water produced. Finally, we converted the moles of water into grams to obtain the final mass of water produced.