QQuestionChemistry
QuestionChemistry
In the reaction $\mathrm{CH}_{4}+ 2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+ 2 \mathrm{H}_{2} \mathrm{O}$ how many moles of oxygen are required to burn 16.0 g of methane?
- 32.0
- 1.00
- 2.00
- 0.500
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Step 1:I'll solve this stoichiometry problem step by step:
Step 2:: Balanced Chemical Equation
\mathrm{CH}_{4} + 2\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}
The balanced equation is already given:
Step 3:: Calculate Molar Mass of Methane (\mathrm{CH}_{4})
- Molar mass of $$\mathrm{CH}_{4}$$ = $$12.01 + 4.032 = 16.042 \mathrm{~g/mol}
- Hydrogen: 1.008 \mathrm{~g/mol} \times 4 = 4.032 \mathrm{~g/mol}
Step 4:: Calculate Moles of Methane
\text{Moles of } \mathrm{CH}_{4} = \frac{16.0 \mathrm{~g}}{16.042 \mathrm{~g/mol}} = 0.997 \mathrm{~mol}
Step 5:: Use Stoichiometric Ratio
From the balanced equation, 1 mole of $$\mathrm{CH}_{4}$$ requires 2 moles of $$\mathrm{O}_{2}
\text{Moles of } \mathrm{O}_{2} = 0.997 \mathrm{~mol} \times 2 = 1.994 \mathrm{~mol} \approx 2.00 \mathrm{~mol}
Final Answer
2.00 moles of oxygen are required to burn 16.0 g of methane.
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