Answer: 233 L of [tex]CO_2[/tex] will be produced from 150 grams of [tex]C_4H_{10}[/tex]
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex] [tex]\text{Moles of} C_4H_{10}=\frac{150g}{58g/mol}=2.59moles[/tex]
The balanced chemical equation is:
[tex]2C_4H_{10}+13O_2(g)\rightarrow 8CO_2+10H_2O[/tex]
According to stoichiometry :
2 moles of [tex]C_4H_{10}[/tex] produce = 8 moles of [tex]CO_2[/tex]
Thus 2.59 moles of [tex]C_4H_{10}[/tex] will produce=[tex]\frac{8}{2}\times 2.59=10.4moles[/tex] of [tex]CO_2[/tex]
Volume of [tex]CO_2=moles\times {\text {Molar volume}}=10.4moles\times 22.4mol/L=233L[/tex]
Thus 233 L of [tex]CO_2[/tex] will be produced from 150 grams of [tex]C_4H_{10}[/tex]
