Question

Use the balanced equation given below to solve the following problem; Calculate the volume in liters of CO produced by the reaction of 175 g of Sb2O3.
Sb2O3 + 3 C --> 2 Sb + 3 CO

40.3 L CO
7.81 L CO
20.2 L CO
13.4 L CO

Answer 1

Answer: 40.3 L

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Sb_2O_3=\frac{175g}{291.5g/mol}=0.600moles$

$Sb_2O_3+3C\rightarrow 2Sb+3CO$  

According to stoichiometry :

1 moles of $Sb_2O_3$ produces = 3 moles of $CO$

Thus 0.600 moles of $Sb_2O_3$ will produce=$\frac{3}{1}\times 0.600=1.80moles$  of $CO$

Volume of $CO=moles\times {\text {Molar volume}}=1.80moles\times 22.4L/mol=40.3L$

Thus 40.3 L of CO is produced.