# How do you find the mass of CO2 produced in a reaction of 150.0g of C6H12 in sufficient (excess) oxygen if the reaction has a 35.00% yield?

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(3) marks are available for this question, so I'm guessing that there are about three steps to this question...?

(3) marks are available for this question, so I'm guessing that there are about three steps to this question...?

This formula indicates that

However, the actual yield of CO2 is 35%.

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To find the mass of CO2 produced, calculate the theoretical yield using the balanced chemical equation. Then, multiply the theoretical yield by the given yield percentage.

Balanced equation: C6H12 + 9O2 → 6CO2 + 6H2O

Molar mass of C6H12: 84.18 g/mol Molar mass of CO2: 44.01 g/mol

Calculate moles of C6H12, then moles of CO2, and finally the mass of CO2:

- Moles of C6H12 = 150.0 g / 84.18 g/mol
- Moles of CO2 (using the mole ratio from the balanced equation)
- Mass of CO2 = Moles of CO2 * Molar mass of CO2

Finally, apply the given yield percentage to find the actual mass of CO2 produced.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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