Which of the following `is//are` correct.
The following reaction occurs: ltrgt `CS_(2) + 3Cl_(2) overset(Delta)rarr C Cl_(4) + S_(2) Cl_(2)`
`1.0 g` of `CS_(2)` and `2.0 g` of `Cl_(2)` reacts.

Options

A. `0.714 g CS_(2)` is used in the reaction.
B. `0.286 g CS_(2)` is in formed.
C. `1.45 g of C Cl_(4)` is formed
D. `0.8 g Cl_(2)` is in excess

Answer

A, B, C

Explanation

`(Mw "of" CS_(2) = 76, Mw "of" Cl_(2) = 71, Mw "of" C Cl_(4) = 154 g mol^(-1))`
Weight of `Cl_(2)` needed
`=(1.0 g CS_(2)) ((1 "mol" CS_(2))/(76 g CS_(2)))((3 "mol" CS_(2))/("mol"CS_(2)))((71 g Cl_(2))/("mol" Cl_(2)))`
`= (1 xx 3 xx 71)/(76) = 2.8 g Cl_(2)` needed
Since there is `2.0 g Cl_(2)` is the limiting quantity
a. Weight of `Cs_(2)` used
`= (2.0 g Cl_(2))((1 "mol" Cl_(2))/(71 g Cl_(2))) ((1 "mol" CS_(2))/(3 "mol" Cl_(2)))((76 g CS_(2))/("mol" CS_(2)))`
`= (2 xx 1 xx 1 xx 76)/(71 xx 3) = 0.714 g CS_(2)` used.
b. Weight of `CS_(2)` excess or formed
`= (1.0 g CS_(2)` present) - `(0.714 g` used)
`= 0.286 g CS_(2)` formed
c. Weight of `C Cl_(4)` formed
`= (2.0 g Cl_(2)) ((1 "mol" Cl_(2))/(71 g Cl_(2)))((1"mol" CCl_(4))/(2"mol"Cl_(2)))`
`((154 g C Cl_(4))/("mol" C Cl_(4)))`
`= (2 xx 1 xx 154)/(71 xx 3) = 1.45 g C Cl_(4)`
d. Wrong