Question

How many grams of oxygen is required for complete combustion of 29g of butane as per the equation

C4H10+ 4.5O2=2CO2+5H2O

Answer 1

atomic weight of C4H10  = 58g
no. of mole of C4H10 = 29/58 = 1/2 mole
for 1 mole of butane combustion 9/2 mole of oxygen is required
for 1/2 mole of butane combustion 9/4 mole of oxygen is required
weight of oxygen = 9*16/4 = 36 gm.

Answer 2

36 grams of Oxygen is required to completely combust 29 grams of butane.

The atomic weight of the compound of butane (C4H10) = atomic weight of Carbon*number of carbon atoms + atomic weight of hydrogen*number of hydrogen atoms

         = 12*4 + 1*10

        = 58 grams

Therefore, 58 grams of butane will contain 1 mole of atoms. Hence, 29 grams of butane contain 0.5 moles.

According to the chemical reaction: C4H10+ 4.5O2=2CO2+5H2O,

To combust 1 one of butane 4.5 moles of oxygen are required. Hence to combust 0.5 moles of Butane, 2.25 moles of oxygen will be required.

One mole of oxygen = 16 grams

For 2.25 moles = 16*2.25 = 36 grams

Therefore, the weight of oxygen required to combust 29 grams of Butane is 36 grams.