QuestionSeptember 20, 2025

When 24 g of methane gas (CH_(4)) is combined with 16 g of oxygen gas, if 15 g of water is collected and the only other product is carbon dioxide, how much carbon dioxide must have been released? 25 a 2.09 40 g 15g

When 24 g of methane gas (CH_(4)) is combined with 16 g of oxygen gas, if 15 g of water is collected and the only other product is carbon dioxide, how much carbon dioxide must have been released? 25 a 2.09 40 g 15g
When 24 g of methane gas (CH_(4)) is combined with 16 g of oxygen gas, if 15 g of water is collected and the only other product is carbon dioxide, how much carbon dioxide must have been released? 25 a 2.09 40 g 15g

Solution
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Answer

18.3 g Explanation 1. Write the balanced chemical equation CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O 2. Calculate moles of reactants Moles of CH_4 = \frac{24}{16} = 1.5 mol; Moles of O_2 = \frac{16}{32} = 0.5 mol. 3. Identify limiting reagent CH_4 needs 2 \times 1.5 = 3 mol O_2, but only 0.5 mol O_2 is present, so O_2 is limiting. 4. Find moles of water produced from limiting reagent From 0.5 mol O_2: \frac{2\,mol\,H_2O}{2\,mol\,O_2} \times 0.5\,mol\,O_2 = 0.5 mol H_2O. 5. Confirm mass of water matches collected amount Mass of 0.5 mol H_2O = 0.5 \times 18 = 9 g (but 15 g collected, so check calculation). 6. Recalculate using actual water collected Moles of H_2O collected = \frac{15}{18} = 0.833 mol. 7. Use stoichiometry to find moles of CO_2 produced From equation, 1 mol CO_2 per 2 mol H_2O. So, moles of CO_2 = \frac{0.833}{2} = 0.416 mol. 8. Calculate mass of CO_2 released Mass = 0.416 \times 44 = 18.3 g.

Explanation

1. Write the balanced chemical equation<br /> $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$<br />2. Calculate moles of reactants<br /> Moles of $CH_4 = \frac{24}{16} = 1.5$ mol; Moles of $O_2 = \frac{16}{32} = 0.5$ mol.<br />3. Identify limiting reagent<br /> $CH_4$ needs $2 \times 1.5 = 3$ mol $O_2$, but only $0.5$ mol $O_2$ is present, so $O_2$ is limiting.<br />4. Find moles of water produced from limiting reagent<br /> From $0.5$ mol $O_2$: $\frac{2\,mol\,H_2O}{2\,mol\,O_2} \times 0.5\,mol\,O_2 = 0.5$ mol $H_2O$.<br />5. Confirm mass of water matches collected amount<br /> Mass of $0.5$ mol $H_2O = 0.5 \times 18 = 9$ g (but 15 g collected, so check calculation).<br />6. Recalculate using actual water collected<br /> Moles of $H_2O$ collected $= \frac{15}{18} = 0.833$ mol.<br />7. Use stoichiometry to find moles of $CO_2$ produced<br /> From equation, $1$ mol $CO_2$ per $2$ mol $H_2O$. So, moles of $CO_2 = \frac{0.833}{2} = 0.416$ mol.<br />8. Calculate mass of $CO_2$ released<br /> Mass $= 0.416 \times 44 = 18.3$ g.
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