Answer: The amount of carbon dioxide formed in the reaction is 0.566 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .......(1)
Mass of oxygen gas = 8.00 grams
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of oxygen gas}=\frac{8g}{32g/mol}=0.25mol[/tex]
For the given chemical equation:
[tex]7O_2(g)+2C_5H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)[/tex]
As, ethane is present in excess. It is considered as an excess reagent.
So, oxygen gas is considered as a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
7 moles of oxygen gas produces 4 moles of carbon dioxide
So, 0.25 moles of oxygen gas will produce = [tex]\frac{4}{7}\times 0.25=0.143moles[/tex] of carbon dioxide
Now, calculating the mass of carbon dioxide from equation 1, we get:
Molar mass of carbon dioxide = 44 g/mol
Moles of carbon dioxide = 0.143 moles
Putting values in above equation, we get:
[tex]0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.143mol\times 44g/mol)=6.29g[/tex]
To calculate the actual yield of carbon dioxide, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100[/tex]
Percentage yield of carbon dioxide = 90 %
Theoretical yield of carbon dioxide = 6.29 g
Putting values in above equation, we get:
[tex]90=\frac{\text{Actual yield of carbon dioxide}}{6.29g}\times 100\\\\\text{Actual yield of carbon dioxide}=\frac{6.29\times 90}{100}=0.566g[/tex]
Hence, the amount of carbon dioxide formed in the reaction is 0.566 grams
