Question #36710

How many grams of O2(g) are needed to completely burn 21.8 g of C3H8(g)?

Expert's answer

The complete burning of propane can be described by following equation:

C3H8 + 5 O2 = 3 CO2 + 4 H2O

The quantity of propane is

n(C3H8) = m(C3H8) / M(C3H8) = 21,8 g / 44 g/mol = 0.495 mol, where

m(C3H8) is the mass of propane,

M(C3H8) is the molar mass of propane, M(C3H8) = 3*12 g/mol + 8*1 g/mol = 44 g/mol.

the needed quantity of oxygen is

n(O2) = 5*n(CO2) / 3 = 5*0.495 mol / 3 = 0.825 mol

the mass of oxygen needed to completely burn of propane is

m(O2) = n(O2) * M(O2) = 0.825 mol * 32 g/mol = 26.4 g.

Answer: 26.4 g of O2(g) are needed to completely burn 21.8 g of C3H8(g).

C3H8 + 5 O2 = 3 CO2 + 4 H2O

The quantity of propane is

n(C3H8) = m(C3H8) / M(C3H8) = 21,8 g / 44 g/mol = 0.495 mol, where

m(C3H8) is the mass of propane,

M(C3H8) is the molar mass of propane, M(C3H8) = 3*12 g/mol + 8*1 g/mol = 44 g/mol.

the needed quantity of oxygen is

n(O2) = 5*n(CO2) / 3 = 5*0.495 mol / 3 = 0.825 mol

the mass of oxygen needed to completely burn of propane is

m(O2) = n(O2) * M(O2) = 0.825 mol * 32 g/mol = 26.4 g.

Answer: 26.4 g of O2(g) are needed to completely burn 21.8 g of C3H8(g).

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