Question

Consider the reaction
C6H12 + 9 O2 –> 6 CO2 + 6 H2O
How many grams of O2 are consumed when C6H12 is burned to yield 2.5 grams of CO2?

Give your answer in 1 decimal.

It said this when I got the wrong answer: Calculate the number of moles of CO2 produced in the reaction. Then match the number of moles of O2 and the corresponding number of grams of O2. Do not forget the reaction stoichiometry.
The correct answer is: 2.7

But I still dont know how to get the answer.

Answer 1

Answer on Question #43459 - Chemistry - Inorganic Chemistry

Question:

Consider the reaction

$$\mathrm{C_6H_{12} + 9O_2 \rightarrow 6CO_2 + 6H_2O}$$

How many grams of $\mathrm{O_2}$ are consumed when $\mathrm{C_6H_{12}}$ is burned to yield 2.5 grams of $\mathrm{CO_2}$?

Give your answer in 1 decimal.

Solution:

Number of moles of $\mathrm{CO_2}$ is calculated as

$$n_{\mathrm{CO_2}} = \frac{m_{\mathrm{CO_2}}}{M_{\mathrm{CO_2}}} = \frac{2.5\,g}{44.0\,g/mol} = 0.057\,mol$$

where $M_{\mathrm{CO_2}}$ – molar weight of $\mathrm{CO_2}$.

As is clear from the reaction stoichiometry, 9 moles of $\mathrm{O_2}$ are consumed to yield 6 moles of $\mathrm{CO_2}$. Having calculated the actual number of moles of $\mathrm{CO_2}$ we can write down the proportion:

$$6\,\mathrm{mol}\,(\mathrm{CO_2}) - 9\,\mathrm{mol}\,(\mathrm{O_2})$$ $$0.057\,\mathrm{mol}\,(\mathrm{CO_2}) - n_{\mathrm{O_2}}\,\mathrm{mol}\,(\mathrm{O_2}),$$

whence

$$n_{\mathrm{O_2}} = \frac{0.057 \cdot 9}{6} = 0.085\,mol$$

Mass of $\mathrm{O_2}$ consumed is

$$m_{\mathrm{O_2}} = n_{\mathrm{O_2}} \cdot M_{\mathrm{O_2}} = 0.085 \cdot 32.0 = 2.7\,g$$

where $M_{\mathrm{O_2}}$ – molar weight of $\mathrm{O_2}$.

Answer: $2.7\,g$

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