Answer :
Absolutely, let's go through the detailed steps for solving this problem:
1. Identify the given data:
- Mass of methane ([tex]\( \text{CH}_4 \)[/tex]): [tex]\( 100 \)[/tex] grams.
- Molar mass of methane ([tex]\( \text{CH}_4 \)[/tex]): [tex]\( 16.04 \)[/tex] grams/mole.
- Molar mass of oxygen ([tex]\( \text{O}_2 \)[/tex]): [tex]\( 32.00 \)[/tex] grams/mole.
- Avogadro's number: [tex]\( 6.022 \times 10^{23} \)[/tex] molecules/mole.
2. Calculate the moles of methane ([tex]\( \text{CH}_4 \)[/tex]) from the given mass:
- The formula to find the number of moles is:
[tex]\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \][/tex]
- For methane:
[tex]\[ \text{moles of } \text{CH}_4 = \frac{100 \text{ g}}{16.04 \text{ g/mol}} \approx 6.234 \][/tex]
Therefore, the moles of methane ([tex]\( \text{CH}_4 \)[/tex]) is approximately [tex]\( 6.234 \)[/tex].
3. Determine the moles of oxygen ([tex]\( \text{O}_2 \)[/tex]) required for the reaction:
- From the stoichiometry of the balanced chemical equation, 1 mole of [tex]\( \text{CH}_4 \)[/tex] reacts with 2 moles of [tex]\( \text{O}_2 \)[/tex]. Hence, the moles of oxygen required is twice the moles of methane.
[tex]\[ \text{moles of } \text{O}_2 = 2 \times \text{moles of } \text{CH}_4 \][/tex]
- Using the calculated moles of methane:
[tex]\[ \text{moles of } \text{O}_2 = 2 \times 6.234 = 12.468 \][/tex]
4. Calculate the number of molecules of oxygen ([tex]\( \text{O}_2 \)[/tex]) consumed:
- Utilizing Avogadro's number [tex]\( (6.022 \times 10^{23} \, \text{molecules/mole}) \)[/tex], the number of molecules can be found by:
[tex]\[ \text{molecules of } \text{O}_2 = \text{moles of } \text{O}_2 \times \text{Avogadro's number} \][/tex]
- For the calculated moles of oxygen:
[tex]\[ \text{molecules of } \text{O}_2 = 12.468 \times 6.022 \times 10^{23} \approx 7.509 \times 10^{24} \][/tex]
Therefore, the number of molecules of oxygen ([tex]\( \text{O}_2 \)[/tex]) consumed in the combustion of [tex]\( 100 \,\text{g} \)[/tex] of methane is approximately [tex]\( 7.509 \times 10^{24} \)[/tex].
To sum up,
- The moles of methane ([tex]\( \text{CH}_4 \)[/tex]) is [tex]\( 6.234 \)[/tex].
- The moles of oxygen ([tex]\( \text{O}_2 \)[/tex]) required is [tex]\( 12.468 \)[/tex].
- The number of molecules of oxygen ([tex]\( \text{O}_2 \)[/tex]) consumed is approximately [tex]\( 7.509 \times 10^{24} \)[/tex].
1. Identify the given data:
- Mass of methane ([tex]\( \text{CH}_4 \)[/tex]): [tex]\( 100 \)[/tex] grams.
- Molar mass of methane ([tex]\( \text{CH}_4 \)[/tex]): [tex]\( 16.04 \)[/tex] grams/mole.
- Molar mass of oxygen ([tex]\( \text{O}_2 \)[/tex]): [tex]\( 32.00 \)[/tex] grams/mole.
- Avogadro's number: [tex]\( 6.022 \times 10^{23} \)[/tex] molecules/mole.
2. Calculate the moles of methane ([tex]\( \text{CH}_4 \)[/tex]) from the given mass:
- The formula to find the number of moles is:
[tex]\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \][/tex]
- For methane:
[tex]\[ \text{moles of } \text{CH}_4 = \frac{100 \text{ g}}{16.04 \text{ g/mol}} \approx 6.234 \][/tex]
Therefore, the moles of methane ([tex]\( \text{CH}_4 \)[/tex]) is approximately [tex]\( 6.234 \)[/tex].
3. Determine the moles of oxygen ([tex]\( \text{O}_2 \)[/tex]) required for the reaction:
- From the stoichiometry of the balanced chemical equation, 1 mole of [tex]\( \text{CH}_4 \)[/tex] reacts with 2 moles of [tex]\( \text{O}_2 \)[/tex]. Hence, the moles of oxygen required is twice the moles of methane.
[tex]\[ \text{moles of } \text{O}_2 = 2 \times \text{moles of } \text{CH}_4 \][/tex]
- Using the calculated moles of methane:
[tex]\[ \text{moles of } \text{O}_2 = 2 \times 6.234 = 12.468 \][/tex]
4. Calculate the number of molecules of oxygen ([tex]\( \text{O}_2 \)[/tex]) consumed:
- Utilizing Avogadro's number [tex]\( (6.022 \times 10^{23} \, \text{molecules/mole}) \)[/tex], the number of molecules can be found by:
[tex]\[ \text{molecules of } \text{O}_2 = \text{moles of } \text{O}_2 \times \text{Avogadro's number} \][/tex]
- For the calculated moles of oxygen:
[tex]\[ \text{molecules of } \text{O}_2 = 12.468 \times 6.022 \times 10^{23} \approx 7.509 \times 10^{24} \][/tex]
Therefore, the number of molecules of oxygen ([tex]\( \text{O}_2 \)[/tex]) consumed in the combustion of [tex]\( 100 \,\text{g} \)[/tex] of methane is approximately [tex]\( 7.509 \times 10^{24} \)[/tex].
To sum up,
- The moles of methane ([tex]\( \text{CH}_4 \)[/tex]) is [tex]\( 6.234 \)[/tex].
- The moles of oxygen ([tex]\( \text{O}_2 \)[/tex]) required is [tex]\( 12.468 \)[/tex].
- The number of molecules of oxygen ([tex]\( \text{O}_2 \)[/tex]) consumed is approximately [tex]\( 7.509 \times 10^{24} \)[/tex].
