Question

1. What mass of water is produced from the complete combustion of 3.10×10−3 g of methane?

2. What mass of oxygen is needed for the complete combustion of 3.10×10⁻³ g of methane?

3. Set up two algebraic equations, one expressing the mass of carbon dioxide produced in terms of each reagent and the other expressing the mass of sample burned in terms of each reagent.

What is the mass of methyl alcohol (CH3OH) in the sample?

Answer 1

1. What mass of water is produced from the complete combustion of 3.10×10−3 g of methane?

CH4 + 2O2 --> CO2 + 2H2O

1 mol of methane : 2 mol of water

mol = mass/MW

mass o methane = 3.1*10^-3 g

MW of methane = 16 g/mol

mol = mass/MW = (3.1*10^-3)/(16) =1.94*10^-4 mol of Methane

there is 2 mol of H2O produced per 1 mol of CH4 so: 2*1.94*10^-4 = 3.87*10^-4 mol of Water will be produced

change to mass

mass = mol*MW = (3.87*10^-4 )(18) = 0.006975 grams of H2O

2. What mass of oxygen is needed for the complete combustion of 3.10×10−3 g of methane?

CH4 + 2O2 --> CO2 + 2H2O

convert mass of methane to mol of CH4

mol = mass/MW = (3.1*10^-3)/16 = 1.937*10^-4 mol of CH4

1 mol of CH4 : 2 mol of O2

so

2*1.937*10^-4 = 3.88*10^-4 mol of O2

MW of O2 = 32 g/mol

mass = mol*MW = 3.88*10^-4 mol of O2 * 32 = 0.0124 grams of O2

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