
From the equation $C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O$
i) what volume of carbon dioxide is produced by $200c{m^3}$ of methane?
ii) What volume of Oxygen is required to completely burn $200c{m^3}$ of Methane?
Answer
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Hint: Considering standard temperature and pressure (STP) conditions, we’ll consider the pressure to be 1 atmosphere and temperature to be 273.15K. At STP 1 mole of molecule occupies $22.4L$ of volume. This is the standard values we’ll consider for the given question.
Complete answer:
The given reaction is a combustion reaction. Combustion of hydrocarbons always leads to the formation of Carbon Dioxide and water. According to the balanced chemical reaction, we can infer that one mole of Methane required 2 moles of Oxygen to form 1 mol of Carbon Dioxide and 2 moles of water.
i) Given that we have taken $200c{m^3}$ of methane. 1 mole of methane produces only 1 mole of carbon dioxide. Hence, we can say that:
$1mol{\text{ }}C{H_4} = 1mol{\text{ }}C{O_2}$
We know that one mole of molecule at STP occupies $22.4L$ of volume. We can say that,
$22.4L{\text{ }}C{H_4} = 22.4L{\text{ C}}{{\text{O}}_2}$
$200c{m^3}{\text{ }}C{H_4} = \dfrac{{0.2 \times 22.4}}{{22.4}} = 0.2L = 200c{m^3}$of carbon dioxide.
ii) According to the balanced chemical equation 1mole of methane requires 2 moles of Oxygen. We can say that:
$1mol{\text{ }}C{H_2}(22.4L) = 2mol{\text{ }}{O_2}(2 \times 22.4)$
Therefore, $200c{m^3}$ or $0.2L$ of methane will require $ = \dfrac{{0.2 \times 2 \times 22.4}}{{22.4}} = 0.4L = 400c{m^3}$ of Oxygen.
Note:
According to Avogadro’s hypothesis at constant Pressure and Temperature, all gases contain the same number of particles. Since the number of particles are almost the same, the volume occupied by them will also be the same i.e., $22.4L$ . This volume will change when a gas is heated as the gas expands on heating, as the molecule moves at a faster rate. Also, change in pressure (expansion or compression) changes the volume of the gas. Hence, while making this assumption the Temperature and pressure is considered to be constant at 272.15K and 1 atmosphere respectively.
Complete answer:
The given reaction is a combustion reaction. Combustion of hydrocarbons always leads to the formation of Carbon Dioxide and water. According to the balanced chemical reaction, we can infer that one mole of Methane required 2 moles of Oxygen to form 1 mol of Carbon Dioxide and 2 moles of water.
i) Given that we have taken $200c{m^3}$ of methane. 1 mole of methane produces only 1 mole of carbon dioxide. Hence, we can say that:
$1mol{\text{ }}C{H_4} = 1mol{\text{ }}C{O_2}$
We know that one mole of molecule at STP occupies $22.4L$ of volume. We can say that,
$22.4L{\text{ }}C{H_4} = 22.4L{\text{ C}}{{\text{O}}_2}$
$200c{m^3}{\text{ }}C{H_4} = \dfrac{{0.2 \times 22.4}}{{22.4}} = 0.2L = 200c{m^3}$of carbon dioxide.
ii) According to the balanced chemical equation 1mole of methane requires 2 moles of Oxygen. We can say that:
$1mol{\text{ }}C{H_2}(22.4L) = 2mol{\text{ }}{O_2}(2 \times 22.4)$
Therefore, $200c{m^3}$ or $0.2L$ of methane will require $ = \dfrac{{0.2 \times 2 \times 22.4}}{{22.4}} = 0.4L = 400c{m^3}$ of Oxygen.
Note:
According to Avogadro’s hypothesis at constant Pressure and Temperature, all gases contain the same number of particles. Since the number of particles are almost the same, the volume occupied by them will also be the same i.e., $22.4L$ . This volume will change when a gas is heated as the gas expands on heating, as the molecule moves at a faster rate. Also, change in pressure (expansion or compression) changes the volume of the gas. Hence, while making this assumption the Temperature and pressure is considered to be constant at 272.15K and 1 atmosphere respectively.
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