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Methane [tex]$\left( CH _4, 16.05 \, \text{g/mol} \right)$[/tex] reacts with oxygen to form carbon dioxide [tex]$\left( CO _2, 44.01 \, \text{g/mol} \right)$[/tex] and water [tex]$\left( H _2 O , 18.02 \, \text{g/mol} \right)$[/tex].

Assume that you design a system for converting methane to carbon dioxide and water. To test the efficiency of the system in the laboratory, you burn [tex]$5.00 \, \text{g}$[/tex] of methane. The actual yield is [tex]$6.10 \, \text{g}$[/tex] of water.

What is your percent yield? [tex]$\qquad$[/tex] [tex]$\%$[/tex]



Answer :

GinnyAnswer
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].

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