Certainly! Let's solve the problem step-by-step:
1. Understand the Given Information:
- The balanced chemical equation for the combustion of methane is:
[tex]\[
CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g)
\][/tex]
- We need to find out how many grams of methane ([tex]\(CH_4\)[/tex]) are required to produce 5.6 liters of carbon dioxide ([tex]\(CO_2\)[/tex]) at Standard Temperature and Pressure (STP).
- The molar volume of an ideal gas at STP is 22.4 liters per mole.
2. Determine Moles of CO_2 Produced:
- Using the molar volume of a gas at STP, we convert the volume of [tex]\(CO_2\)[/tex] to moles:
[tex]\[
\text{Moles of } CO_2 = \frac{\text{Volume of } CO_2}{\text{Molar Volume at STP}} = \frac{5.6 \text{ L}}{22.4 \text{ L/mol}} = 0.25 \text{ moles}
\][/tex]
3. Relate Moles of [tex]\(CO_2\)[/tex] to Moles of [tex]\(CH_4\)[/tex]:
- According to the balanced chemical equation, 1 mole of [tex]\(CH_4\)[/tex] produces 1 mole of [tex]\(CO_2\)[/tex].
- Therefore, the moles of [tex]\(CH_4\)[/tex] required to produce 0.25 moles of [tex]\(CO_2\)[/tex] is also 0.25 moles.
4. Calculate the Mass of Methane Required:
- The molar mass of methane ([tex]\(CH_4\)[/tex]) is known to be 16.0 grams per mole.
- The mass of methane needed is given by:
[tex]\[
\text{Mass of } CH_4 = \text{Moles of } CH_4 \times \text{Molar Mass of } CH_4 = 0.25 \text{ moles} \times 16.0 \text{ g/mol} = 4.0 \text{ grams}
\][/tex]
5. Conclusion:
- Therefore, to produce 5.6 liters of carbon dioxide at STP, you need to burn 4.0 grams of methane ([tex]\(CH_4\)[/tex]).
The correct answer is [tex]\( \boxed{4.0 \text{ g}} \)[/tex].
