Question

In an experiment, 170.9 g of C2H4 was reacted with an excess of O2, 164.1 g of CO2 is produced.

Answer 1

Answer:

$Y=30.6\%$

Explanation:

Hello,

In this case, given the reaction, the molar mass of ethene is 28 g/mol and the molar mass of carbon dioxide is 44 g/mol. With that information we compute the theoretical yield considering a 1:2 molar ratio respectively between them:

$m_{CO_20}^{theoretical}=170.9gC_2H_4*\frac{1molC_2H_4}{28gC_2H_4}*\frac{2molCO_2}{1molC_2H_4} *\frac{44gCO_2}{1molCO_2} =537.1gCO_2$

Thus, we compute the percent yield with the given grams of carbon dioxide:

$Y=\frac{m_{CO_2}^{real}}{m_{CO_2}^{theoretical}}*100 \% =\frac{164.1g}{537.1gCO_2} *100 \%\\\\Y=30.6\%$

Regards.