Question

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

C2H4 (g) + 3 O2(g) → 2 CO2 (g) + 2 H2O (l)


What is the percent yield of this reaction?

Answer 1

`Y=30.6\%`

Step-by-step 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*(1molC_2H_4)/(28gC_2H_4)*(2molCO_2)/(1molC_2H_4) *(44gCO_2)/(1molCO_2) =537.1gCO_2`

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

`Y=(m_(CO_2)^(real))/(m_(CO_2)^(theoretical))*100 \% =(164.1g)/(537.1gCO_2) *100 \%\\\\Y=30.6\%`

Regards.