Question

CS2(g) + O2(g) CO2(g) + SO2(g)

Consider the unbalanced equation above. A 0.150 g sample of CS2 is reacted with 285 mL of O2 at STP. What is the total volume of the final mixture of gases at STP?

Answer 1

Given equation is $CS_2+O_2\to CO_2+SO_2$

Balanced equation will be $CS_2+3O_2\to CO_2+2SO_2$

As 1 mole of $CS_2$ reacts with 3 moles of $O_2$ it gives 1 mole of $CO_2$ and 2 moles of $SO_2$

Molar mass of $CS_2$ $=$ 76$g/mol$

Number of moles of $CS_2 =\frac{150}{76}=0.001974$

Number of moles of $O_2$ $=\frac{285}{22400}=0.0127$

Number of moles of $O_2$ required to completely react with $CS_2$ $=3\times0.001974=0.005922$

Number of unreacted moles of $O_2=0.0127-0.005922=0.006778$

Number of moles of $CO_2$ formed$=0.001974$

Number of moles of $SO_2$ formed$=2\times0.001974=0.003948$

After reaction total number of gaseous moles$=0.006778+0.001974+0.003948=0.0127$

Total volume of final mixture $=0.0127\times22400=285ml$