What mass of CH4 is required to produce 14.5 g of H2O if the reaction CH4 + 2O2 -> CO2 + 2H2O has a 73.5% yield?

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The reaction between methane and oxygen to give water and carbon dioxide is CH4 + 2O2 --> CO2 + 2H2O.

It is given that the reaction has a 73.5% yield, and the mass of methane is required that will yield 14.5 g of water.

According to the equation for 1 molecule of methane, 2 molecules of water are produced. Taking the reaction yield into consideration the number of methane molecules required to give one molecule of water is 0.5/0.735

Water has a molar mass of 18 g/mole. 14.5 grams of water is 14.5/18 mole of water. This requires (14.5/18)*(0.5/0.735)*16 = 8.767 g of methane.

The required mass of methane to produce 14.5 g of water is 8.767 g.

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