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?

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:

[tex]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[/tex]

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

[tex]Y=\frac{m_{CO_2}^{real}}{m_{CO_2}^{theoretical}}*100 \% =\frac{164.1g}{537.1gCO_2} *100 \%\\\\Y=30.6\%[/tex]

Regards.

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?

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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?