Answer: 15.0 g of [tex]CCl_4[/tex] will be formed upon the complete reaction of 27.7 grams of chlorine gas with excess carbon disulfide
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Cl_2=\frac{27.7g}{71g/mol}=0.390moles[/tex]
The balanced chemical reaction is:
[tex]CS_2(g)+4Cl_2(g)\rightarrow CCl_4(l)+2SCl_2(s)[/tex]
[tex]Cl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]CS_2[/tex] is the excess reagent.
According to stoichiometry :
4 moles of [tex]Cl_2[/tex] give= 1 mole of [tex]CCl_4[/tex]
Thus 0.390 moles of will give = [tex]\frac{1}{4}\times 0.390=0.0975moles[/tex] of [tex]CCl_4[/tex]
Mass of [tex]CCl_4=moles\times {\text {Molar mass}}=0.0975\times 154=15.0g[/tex]
Thus 15.0 g of [tex]CCl_4[/tex] will be formed upon the complete reaction of 27.7 grams of chlorine gas with excess carbon disulfide
