Chemistry, 26.02.2020 01:33 rouna

Liquid hexane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If of carbon dioxide is produced from the reaction of of hexane and of oxygen gas, calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.

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Answer from: kyaisaambrose6211

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.

The percent yield of carbon dioxide is 75.0 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(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:

$\text{Moles of hexane}=\frac{4.31g}{86.2g/mol}=0.05mol$

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:

$\text{Moles of oxygen gas}=\frac{26.9g}{32g/mol}=0.841mol$

The chemical equation for the combustion of hexane follows:

$2C_6H_{14}(l)+19O_2(g)\rightarrow 12CO_2(g)+14H_2O(g)$

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 = 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 = $\frac{12}{2}\times 0.05=0.3moles$ of carbon dioxide gas

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:

$0.3mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.3mol\times 44g/mol)=13.2g$

To calculate the percentage yield of carbon dioxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of carbon dioxide = 9.90 g

Theoretical yield of carbon dioxide = 13.2 g

Putting values in above equation, we get:

$\%\text{ yield of carbon dioxide}=\frac{9.90g}{13.2g}\times 100\\\\\% \text{yield of carbon dioxide}=75.0\%$

Hence, the percent yield of the carbon dioxide is 75.0 %.

Answer from: Quest

the answer is b & c i believe : )

Answer from: Quest

i think it is true since you can test the answer and can use observations to complete it.

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