The coefficients in a balanced chemical equation provide the mole-to-mole stoich

Question

The coefficients in a balanced chemical equation provide the mole-to-mole stoichiometry among the reactants and products. The molar mass (in g/mol) can be used as the conversion factor between moles and the mass of a substance. Thus, the balanced equation and molar masses can be used in conjunction with one another to calculate the masses involved in a reaction. What coefficients are needed to balance the equation for the complete combustion of methane? Enter the coefficients in the order CH4 . O2. CO2, and H2O, respectively. What mass of carbon dioxide is produced from the complete combustion of 6.10 Times 10-3g of methane? What mass of water is produced from the complete combustion of 6.10 Times10-3g of methane? What mass of oxygen is needed for the complete combustion of 6.10 Times 10-3g of methane?

Explanation / Answer

1. CH4 + 2O2 -----------> CO2 + 2H2O

so answer is 1,2,1,2

PART B

mass of methane = 6.10 * 10^(-3) g , molecular mass of methane = 12+4 = 16

moles = 6.10 * 10^(-3) / 16 = 3.8 * 10^(-4) moles

moles of CO2 produced = moles of methane

mass of CO2 = 3.8 * 10^(-4) * 12 = 4.5 * 10^(-3) g

PART C

moles of water = 2 * moles of methane

=> mass of H2O = 2 * 3.8 * 10^(-4) * 18 = 0.0136 g

PART D

moles of O2 needed = 2* moles of methane

mass of O2 = 2 * 3.8 * 10^(-4) * 32 = 0.024 g

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