Answer:
[tex]Y=75.6\%[/tex]
Explanation:
Hello.
In this case, since no information about the reacting hydrogen is given, we can assume that it completely react with the 28.0 g of acetylene to yield ethane. In such a way, via the 1:1 mole ratio between acetylene (molar mass = 26 g/mol) and ethane (molar mass = 30 g/mol), we compute the yielded grams, or the theoretical yield of ethane as shown below:
[tex]m_{C_2H_6}^{theoretical}=28.0gC_2H_2*\frac{1molC_2H_2}{26gC_2H_2}*\frac{1molC_2H_6}{1molC_2H_2} *\frac{30gC_2H_6}{1molC_2H_6}\\ \\m_{C_2H_6}^{theoretical}=32.3gC_2H_6[/tex]
Hence, by knowing that the percent yield is computed via the actual yield (24.5 g) over the theoretical yield, we obtain:
[tex]Y=\frac{24.5g}{32.3g}*100\%\\ \\Y=75.6\%[/tex]
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