Question

Balanced equation: CH4(g)+2O2(g)→CO2(g)+2H2O(g)

Part B: What mass of carbon dioxide is produced from the complete combustion of 2.70×10⁻³ g of methane?

Part C: What mass of water is produced from the complete combustion of 2.70×10⁻³ g of methane?

PartD: What mass of oxygen is needed for the complete combustion of 2.70×10⁻³ g of methane?

Answer 1

The balanced equation is CH4 (g) + 2 02 (g) -----------> CO2 (g) + 2 H20 (g) Mass of methane = 2.70 x10-3 g = 0.0027 g Number of moles of methane = mass/mol.wt. = 0.0027 g/(16.04 g) = 0.000168 mol Part-B: According to balanced equation, 1 mole of CH4 produces 1 mole of CO2 So, 0.000168 mol of CH4 would produce 0.000168 mol of CO2 Mass of CO2 = moles * mol.wt. = 0.000168 mol * 44.01 g/mol = 7.41x10-3 g Part-C: According to balanced equation, 1 mole of CH4 produces 2 moles of H20 So, 0.000168 mol of CH4 would produce 0.000168*2/1 = 0.000337 mol of H2O Mass of H2O = moles * mol.wt. = 0.000337 mol * 18.015 g/mol = 6.06 *10-3 g Part-D: According to balanced equation, 1 mole of CH4 reacts with 2 moles of O2 So, 0.000168 mol of CH4 would react with 0.000168*2/1 = 0.000337 mol of O, Mass of O, = moles * mol.wt. = 0.000337 mol * 32.00 g/mol = 1.08 x10-2 g