Question

2C4H10+13O2-->8CO2+10H2O Using the predicted and balanced equation, How many Liters of CO2 can be produced from 150 grams of C4H10?

Answer 1

Answer: 233 L of
`CO_2` will be produced from 150 grams of
`C_4H_(10)`

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`
`\text{Moles of} C_4H_(10)=(150g)/(58g/mol)=2.59moles`

The balanced chemical equation is:

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

According to stoichiometry :

2 moles of
`C_4H_(10)` produce = 8 moles of
`CO_2`

Thus 2.59 moles of
`C_4H_(10)` will produce=
`(8)/(2)* 2.59=10.4moles` of
`CO_2`

Volume of
`CO_2=moles* {\text {Molar volume}}=10.4moles* 22.4mol/L=233L`

Thus 233 L of
`CO_2` will be produced from 150 grams of
`C_4H_(10)`