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  $C_4H_{10}$

Explanation:

To calculate the moles :

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

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=$\frac{8}{2}\times 2.59=10.4moles$  of $CO_2$  

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

Thus 233 L of $CO_2$ will be produced from 150 grams of  $C_4H_{10}$