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Excess liquid bexacarbon decahydride reacts with oxygen gas (O2) to produce liquid water (H2O) and carbon

dioxide gas. In the laboratory, a chemist carries out this reaction with 35.0 g of hexacarbon decahydride, and
45.0 grams of oxygen gas and produces 42.6g of carbon dioxide.
Write a balanced equation for the reaction.
Calculate the theoretical yield of carbon dioxide gas.
Calculate the percent yield of the reaction.
(One mole of any gas occupies 22.4 L under certain conditions of temperature and pressure. Assume those
conditions for this question.)

1 Answer

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Sure! Let's tackle this step by step.

To write a balanced equation for the reaction, we need to ensure that the number of atoms on both sides of the equation is equal. Based on the information provided, the balanced equation for the reaction is:

C6H12 + 9O2 -> 6CO2 + 6H2O

Now, let's move on to calculating the theoretical yield of carbon dioxide gas. We can use the balanced equation to determine the stoichiometry of the reaction. From the equation, we can see that 1 mole of hexacarbon decahydride (C6H12) produces 6 moles of carbon dioxide (CO2).

First, we need to convert the given mass of hexacarbon decahydride (35.0 g) to moles. The molar mass of hexacarbon decahydride is 84.18 g/mol.

Moles of C6H12 = Mass of C6H12 / Molar mass of C6H12
Moles of C6H12 = 35.0 g / 84.18 g/mol

Next, we can use the stoichiometry of the balanced equation to determine the moles of carbon dioxide produced.

Moles of CO2 = Moles of C6H12 x (6 moles of CO2 / 1 mole of C6H12)

Now, we can convert the moles of carbon dioxide to grams using the molar mass of carbon dioxide, which is 44.01 g/mol.

Theoretical yield of CO2 = Moles of CO2 x Molar mass of CO2

Finally, we can calculate the percent yield of the reaction using the formula:

Percent yield = (Actual yield / Theoretical yield) x 100

Given that the actual yield of carbon dioxide is 42.6 g, you can substitute the values into the formula to find the percent yield.

I hope this helps! Let me know if you have any further questions.
User Luis Perez
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