Question

How many grams of O2(g) are needed to completely burn 73.7 g of C3H8(g)?

Answer 1

Answer: 267.2 grams

Explanation:

`C_3H_8+5O_2\rightarrow 3CO_2+4H_2O`

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

`\text{Number of moles of propane}=(73.7g)/(44.1g/mol)=1.67moles`

According to stoichiometry:

1 mole of
`C_3H_8` require 5 moles of
`O_2`

Thus 1.67 moles of
`C_3H_8` will require =
`(5)/(1)* 1.67=8.35moles` of
`O_2`

Mass of
`O_2=moles* {\text {Molar mass}}=8.35moles* 32g/mol=267.2`

Thus 267.2 grams of oxygen are required.

Answer 2

The balanced reaction is:

C3H8 + 5O2 = 3CO2 + 4H2O

We are given the amount of C3H8 to be reacted with O2. This will be the starting point of the calculations.

73.7 g C3H8 ( 1 mol C3H8/ 44.1 g C3H8 ) ( 5 mol O2 / 1 mol C3H8 ) ( 32.0 g O2 / 1 mol O2 ) = 267.39 g O2