A mixture of 4.41 g propane (C 3 H 8 ) and 6.40 g of oxygen (O 2 ) is ignited. W

Question

A mixture of 4.41 g propane (C3H8) and 6.40 g of oxygen (O2) is ignited. What reactant is in excess and by how much assuming complete combustion to form carbon dioxide and water.
The balanced equation is:C3H8(g) + 5O2(g) ? 3CO2(g) + 4H2O(l)

Please explain

A mixture of 4.41 g propane (C3H8) and 6.40 g of oxygen (O2) is ignited. What reactant is in excess and by how much assuming complete combustion to form carbon dioxide and water. The balanced equation is:C3H8(g) + 5O2(g) ? 3CO2(g) + 4H2O(l) Please explain

Explanation / Answer

1 mole of C3H8 reacts with 5 moles O2 to produce 3 moles of CO2

44 gm C3H8 reacts with 160 gm O2 to produce 132 gm CO2

1 gm C3H8 requires 3.63 gm Oxygen for complete combustion

4.41 gm needs 16.036 gm O2 for complete combustion

But since only 6.40 gm Oxygen is available,

It reacts with (4.41/16.036)x6.40

=1.76 gm of C3H8

Remaining amount of C3H8 is (4.41-1.76)=2.65gm

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