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Sagot :
To determine the percent yield of water when burning methane using the given chemical reaction, we will follow these steps:
1. Write the balanced chemical equation:
[tex]\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \][/tex]
This equation indicates that 1 mole of methane (CH[tex]\(_4\)[/tex]) reacts with 2 moles of oxygen (O[tex]\(_2\)[/tex]) to produce 1 mole of carbon dioxide (CO[tex]\(_2\)[/tex]) and 2 moles of water (H[tex]\(_2\)[/tex]O).
2. Calculate the moles of methane burned:
The molar mass of methane (CH[tex]\(_4\)[/tex]) is 16.05 g/mol. Given that we burned 5.00 grams of methane, we calculate the moles of methane as:
[tex]\[ \text{Moles of methane} = \frac{\text{mass of methane}}{\text{molar mass of methane}} = \frac{5.00 \, \text{g}}{16.05 \, \text{g/mol}} = 0.3115 \, \text{moles (rounded to four decimal places)} \][/tex]
3. Calculate the theoretical yield of water in moles:
According to the balanced equation, 1 mole of methane produces 2 moles of water. Therefore, the moles of water produced from 0.3115 moles of methane are:
[tex]\[ \text{Theoretical moles of water} = 0.3115 \, \text{moles of CH}_4 \times 2 = 0.6231 \, \text{moles (rounded to four decimal places)} \][/tex]
4. Convert the theoretical yield of water to grams:
The molar mass of water (H[tex]\(_2\)[/tex]O) is 18.02 g/mol. Therefore, we calculate the theoretical yield in grams as:
[tex]\[ \text{Theoretical yield of water} = 0.6231 \, \text{moles} \times 18.02 \, \text{g/mol} = 11.2274 \, \text{grams (rounded to four decimal places)} \][/tex]
5. Calculate the percent yield:
The actual yield of water obtained from the experiment is 6.10 grams. The percent yield is calculated as:
[tex]\[ \text{Percent yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100 = \left( \frac{6.10 \, \text{g}}{11.2274 \, \text{g}} \right) \times 100 \approx 54.33\% \][/tex]
Therefore, the percent yield of water when burning 5.00 grams of methane is approximately [tex]\(54.33\%\)[/tex].
1. Write the balanced chemical equation:
[tex]\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \][/tex]
This equation indicates that 1 mole of methane (CH[tex]\(_4\)[/tex]) reacts with 2 moles of oxygen (O[tex]\(_2\)[/tex]) to produce 1 mole of carbon dioxide (CO[tex]\(_2\)[/tex]) and 2 moles of water (H[tex]\(_2\)[/tex]O).
2. Calculate the moles of methane burned:
The molar mass of methane (CH[tex]\(_4\)[/tex]) is 16.05 g/mol. Given that we burned 5.00 grams of methane, we calculate the moles of methane as:
[tex]\[ \text{Moles of methane} = \frac{\text{mass of methane}}{\text{molar mass of methane}} = \frac{5.00 \, \text{g}}{16.05 \, \text{g/mol}} = 0.3115 \, \text{moles (rounded to four decimal places)} \][/tex]
3. Calculate the theoretical yield of water in moles:
According to the balanced equation, 1 mole of methane produces 2 moles of water. Therefore, the moles of water produced from 0.3115 moles of methane are:
[tex]\[ \text{Theoretical moles of water} = 0.3115 \, \text{moles of CH}_4 \times 2 = 0.6231 \, \text{moles (rounded to four decimal places)} \][/tex]
4. Convert the theoretical yield of water to grams:
The molar mass of water (H[tex]\(_2\)[/tex]O) is 18.02 g/mol. Therefore, we calculate the theoretical yield in grams as:
[tex]\[ \text{Theoretical yield of water} = 0.6231 \, \text{moles} \times 18.02 \, \text{g/mol} = 11.2274 \, \text{grams (rounded to four decimal places)} \][/tex]
5. Calculate the percent yield:
The actual yield of water obtained from the experiment is 6.10 grams. The percent yield is calculated as:
[tex]\[ \text{Percent yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100 = \left( \frac{6.10 \, \text{g}}{11.2274 \, \text{g}} \right) \times 100 \approx 54.33\% \][/tex]
Therefore, the percent yield of water when burning 5.00 grams of methane is approximately [tex]\(54.33\%\)[/tex].
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