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What mass of oxygen is required for complete combustion of 128g of methane?

CH4+2O2 → CO2+2H2O

User Epieters
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1 Answer

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The balanced chemical equation for the combustion of methane is:

CH4 + 2O2 → CO2 + 2H2O

From the equation, we can see that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.

The molar mass of methane (CH4) is:

(1 x 12.01 g/mol) + (4 x 1.01 g/mol) = 16.05 g/mol

To calculate the amount of oxygen required for the combustion of 128g of methane, we need to first convert the mass of methane to moles:

128 g CH4 ÷ 16.05 g/mol = 7.98 mol CH4

According to the stoichiometry of the balanced equation, 2 moles of oxygen are required for every 1 mole of methane that reacts. Therefore, we need:

7.98 mol CH4 x (2 mol O2 / 1 mol CH4) = 15.96 mol O2

Finally, we can convert the moles of oxygen to grams:

15.96 mol O2 x 32.00 g/mol = 511.2 g O2

Therefore, 511.2 g of oxygen are required for the complete combustion of 128g of methane.

User Roma Rush
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