Question

If this butane lighter uses 15.0 grams of butane, how many grams of carbon dioxide does it produce at 95% efficiency?

Answer 1

Answer: 43.5g of carbon dioxide it produce at 95% efficiency

Step-by-step explanation:

`2C_4H_(10)+13O_2\rightarrow 8CO_2+10H_2O`

To calculate the moles, we use the equation:

moles of butane =
`\frac{\text {given mass}}{\text {molar mass}}=(15.0g)/(58g/mol)=0.26moles`

According to stoichiometry:

2 moles of butane gives = 8 moles of carbon dioxide

Thus 0.26 moles of butane gives=
`(8)/(2)* 0.26=1.04` moles of carbon dioxide

Mass of
`CO_2` produced=
`moles* {\text {Molar mass}}=1.04* 44=45.8g`

As yield of the reaction is 95%, the amount of
`CO_2` produced =
`(95)/(100)* 45.8=43.5g`

Thus 43.5g of carbon dioxide it produce at 95% efficiency