Question

When 34.4 g of carbon is heated with silicon dioxide, 26.9 g of carbon monoxide is produced. What is the percent yield of carbon monoxide for this reaction? SiO2(s) 3C(s)⟶ΔSiC(s) 2CO(g)

Answer 1

Answer:

To calculate the percent yield of carbon monoxide (CO) for this reaction, we need to compare the actual yield with the theoretical yield.

First, let's calculate the theoretical yield of carbon monoxide (CO) using the given information:

Molar mass of carbon monoxide (CO) = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol

Molar mass of carbon (C) = 12.01 g/mol

According to the balanced equation, the ratio between carbon monoxide (CO) and carbon (C) is 2:3.

Therefore, the theoretical yield of carbon monoxide (CO) can be calculated as follows:

Theoretical yield of CO = (mass of C / molar mass of C) * (molar mass of CO / 3)

Theoretical yield of CO = (26.9 g / 12.01 g/mol) * (28.01 g/mol / 3) = 62.02 g

Now, let's calculate the percent yield using the actual yield and theoretical yield:

Percent yield = (actual yield / theoretical yield) * 100

Percent yield = (26.9 g / 62.02 g) * 100 = 43.36%

Therefore, the percent yield of carbon monoxide (CO) for this reaction is approximately 43.36%.

Explanation:

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