Respuesta :
The question is incomplete, here is the complete question:
Liquid hexane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 9.90 g of carbon dioxide is produced from the reaction of 4.31 g of hexane and 26.9 g of oxygen gas, calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.
Answer: The percent yield of carbon dioxide is 75.0 %
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] Β Β Β .....(1)
- For hexane:
Given mass of hexane = 4.31 g
Molar mass of hexane = 86.2 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of hexane}=\frac{4.31g}{86.2g/mol}=0.05mol[/tex]
- For oxygen gas:
Given mass of oxygen gas = 26.9 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of oxygen gas}=\frac{26.9g}{32g/mol}=0.841mol[/tex]
The chemical equation for the combustion of hexane follows:
[tex]2C_6H_{14}(l)+19O_2(g)\rightarrow 12CO_2(g)+14H_2O(g)[/tex]
By Stoichiometry of the reaction:
2 moles of hexane reacts with 19 moles of oxygen gas
So, 0.05 moles of hexane will react with = Β of oxygen gas
As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.
Thus, hexane is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of hexane produces 12 moles of carbon dioxide gas
So, 0.05 moles of hexane will produce = [tex]\frac{12}{2}\times 0.05=0.3moles[/tex] of carbon dioxide gas Β
Now, calculating the mass of carbon dioxide from equation 1, we get:
Molar mass of carbon dioxide = 44 g/mol
Moles of carbon dioxide = 0.3 moles
Putting values in equation 1, we get:
[tex]0.3mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.3mol\times 44g/mol)=13.2g[/tex]
- To calculate the percentage yield of carbon dioxide, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of carbon dioxide = 9.90 g
Theoretical yield of carbon dioxide = 13.2 g
Putting values in above equation, we get:
[tex]\%\text{ yield of carbon dioxide}=\frac{9.90g}{13.2g}\times 100\\\\\% \text{yield of carbon dioxide}=75.0\%[/tex]
Hence, the percent yield of the carbon dioxide is 75.0 %.
