Answer :
To find the percent yield of water from the combustion of methane in the reaction:
[tex]\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \][/tex]
We need to follow these steps:
1. Calculate the moles of methane that were burned:
The molar mass of methane (CH₄) is given as [tex]\( 16.05 \text{ g/mol} \)[/tex].
Given mass of methane burned is [tex]\( 5.00 \text{ g} \)[/tex].
Moles of methane burned:
[tex]\[ \text{moles of } CH_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{5.00 \text{ g}}{16.05 \text{ g/mol}} = 0.311526 \text{ mol} \][/tex]
2. Determine the theoretical moles of water produced:
The balanced chemical equation shows that 1 mole of CH₄ produces 2 moles of H₂O.
From the moles of methane burned:
[tex]\[ \text{theoretical moles of } H_2O = 2 \times \text{moles of } CH_4 = 2 \times 0.311526 \text{ mol} = 0.623053 \text{ mol} \][/tex]
3. Calculate the theoretical mass of water produced:
The molar mass of water (H₂O) is given as [tex]\( 18.02 \text{ g/mol} \)[/tex].
Using the theoretical moles of H₂O determined:
[tex]\[ \text{theoretical mass of } H_2O = \text{theoretical moles of } H_2O \times \text{molar mass of } H_2O \][/tex]
[tex]\[ \text{theoretical mass of } H_2O = 0.623053 \text{ mol} \times 18.02 \text{ g/mol} = 11.2274 \text{ g} \][/tex]
4. Determine the percent yield:
Percent yield is given by the ratio of the actual yield to the theoretical yield, multiplied by 100%.
Given the actual yield of water is [tex]\( 6.10 \text{ g} \)[/tex]:
[tex]\[ \text{percent yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\% \][/tex]
[tex]\[ \text{percent yield} = \left( \frac{6.10 \text{ g}}{11.2274 \text{ g}} \right) \times 100\% = 54.3313\% \][/tex]
Thus, the percent yield of the water produced in this reaction is approximately [tex]\( 54.33\% \)[/tex].
[tex]\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \][/tex]
We need to follow these steps:
1. Calculate the moles of methane that were burned:
The molar mass of methane (CH₄) is given as [tex]\( 16.05 \text{ g/mol} \)[/tex].
Given mass of methane burned is [tex]\( 5.00 \text{ g} \)[/tex].
Moles of methane burned:
[tex]\[ \text{moles of } CH_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{5.00 \text{ g}}{16.05 \text{ g/mol}} = 0.311526 \text{ mol} \][/tex]
2. Determine the theoretical moles of water produced:
The balanced chemical equation shows that 1 mole of CH₄ produces 2 moles of H₂O.
From the moles of methane burned:
[tex]\[ \text{theoretical moles of } H_2O = 2 \times \text{moles of } CH_4 = 2 \times 0.311526 \text{ mol} = 0.623053 \text{ mol} \][/tex]
3. Calculate the theoretical mass of water produced:
The molar mass of water (H₂O) is given as [tex]\( 18.02 \text{ g/mol} \)[/tex].
Using the theoretical moles of H₂O determined:
[tex]\[ \text{theoretical mass of } H_2O = \text{theoretical moles of } H_2O \times \text{molar mass of } H_2O \][/tex]
[tex]\[ \text{theoretical mass of } H_2O = 0.623053 \text{ mol} \times 18.02 \text{ g/mol} = 11.2274 \text{ g} \][/tex]
4. Determine the percent yield:
Percent yield is given by the ratio of the actual yield to the theoretical yield, multiplied by 100%.
Given the actual yield of water is [tex]\( 6.10 \text{ g} \)[/tex]:
[tex]\[ \text{percent yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\% \][/tex]
[tex]\[ \text{percent yield} = \left( \frac{6.10 \text{ g}}{11.2274 \text{ g}} \right) \times 100\% = 54.3313\% \][/tex]
Thus, the percent yield of the water produced in this reaction is approximately [tex]\( 54.33\% \)[/tex].
