What mass of oxygen is needed for the complete combustion of 8.20×10−3g of methane?
What mass of water is produced from the complete combustion of 8.20×10−3g of methane?
Update:What mass of carbon dioxide is produced from the complete combustion of 8.20×10−3g of methane?
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Balanced equation:
CH4 + 2O2 → CO2 + 2H2O
1mol methane reacts with 2 mol O2 to produce 1 mol CO2 and 2 mol H2O
Molar mass CH4 = 16g/mol
Moles in (8.2*10^-3g) methane = (8.2*10^-3) / 16 = 0.0005125 mol CH4
Substitute into word equation:
5.12*10^-4 mol CH4 will react with 1.025*10^-3 mol O2 to produce 5.125*10^-4 mol CO2 and 1.025*10^-3 mol H2O
Calculate masses :
O2 molar mass = 32 g/mol
Mass of 1.025*10^-3 mol = 1.025*10^-3 *32 = 3.28*10^-2g O2
CO2 Molar mass = 44g/mol
Mass of 5.12*10^-4 mol = 5.12*10^-4 * 44 = 2.25*10^-2 g CO2 produced
H2O Molar mass = 18g/mol
Mass of 1.025*10^-3 mol = 1.025*10^-3 * 18 = 1.85*10^-2 g H2O
Does this balance out :
Mass of reactants ( CH4 + O2) = (8.20*10^-3 + 3.28*10^-2) = 4.10*10^-2 g reactants
Mass of products ( CO2 + H2O) = (2.25*10^-2 + 1.85*10^-2 ) = 4.10*10^-2g products
Mass reactants matches mass of products.
CH4+2O2âCO2+2H2O
According to equation 1mol of CH4 need 2mol of O2.
Moles of CH4=8.2Ã10-3/16=5.125Ã10-4mol
So 1.025Ã10-3mol of O2 is needed.
Mass of O2=1.025Ã10-3Ã16=16.4Ã10-3 g
64 gm carbon + 64 gm o2 = 44 gm co2 explain
CH4 + 2O2 ------------> CO2 + 2H2O
16gm --64gm-----------44gm----36gm
for combustion of 16 gm the methane oxygen required = 64 gm
---------------------8.2x10^-3gm-------------------------------------= 64x8.2x10^-3/16
= 32.8x10^-3gm O2 ANSWER
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WATER PRODUCED = [36X32,8X10^-3X8,2X10^-3]/16X64
= 9.45 x10^-6 gm ANSWER