When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 18 grams of water from 8 grams of methane and 32 grams of oxygen, how many grams of carbon dioxide were produced in the reaction? 40 22 58 18 none of the above
Question
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 18 grams of water from 8 grams of methane and 32 grams of oxygen, how many grams of carbon dioxide were produced in the reaction? 40 22 58 18 none of the above
Solution
The balanced chemical equation for the combustion of methane is:
CH4 + 2O2 -> CO2 + 2H2O
From this equation, we can see that one molecule of methane (CH4) reacts with two molecules of oxygen (O2) to produce one molecule of carbon dioxide (CO2) and two molecules of water (H2O).
The molar mass of methane (CH4) is approximately 16 grams/mole, and the molar mass of water (H2O) is approximately 18 grams/mole.
Given that 8 grams of methane were used, this corresponds to 8/16 = 0.5 moles of methane.
Since the reaction produces 2 moles of water for every mole of methane, this means that 0.5 * 2 = 1 mole of water was produced.
Given that 18 grams of water were produced, this corresponds to 18/18 = 1 mole of water, which matches our calculation.
Since the reaction produces 1 mole of carbon dioxide for every mole of methane, this means that 0.5 * 1 = 0.5 moles of carbon dioxide were produced.
The molar mass of carbon dioxide (CO2) is approximately 44 grams/mole, so 0.5 moles of carbon dioxide corresponds to 0.5 * 44 = 22 grams of carbon dioxide.
Therefore, the answer is 22 grams of carbon dioxide.
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