Question

The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g) How many grams of O 2 ( g ) O2(g) are needed to completely burn 19.7 g C 3 H 8 ( g ) ?

Answer 1

Answer: 72 grams of
`O_2(g)` are needed to completely burn 19.7 g
`C_3H_8(g)`

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number
`6.023* 10^(23)` of particles.

To calculate the number of moles, we use the equation:

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

Putting in the values we get:

`\text{Number of moles}=(19.7g)/(44g/mol)=0.45moles`

`C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)`

According to stoichiometry:

1 mole of
`C_3H_8` requires 5 moles of oxygen

0.45 moles of
`C_3H_8` require=
`(5)/(1)* 0.45=2.25` moles of oxygen

Mass of
`O_2=moles* {\text {Molar mass}}=2.25* 32=72g`

72 grams of
`O_2(g)` are needed to completely burn 19.7 g
`C_3H_8(g)`