Answer:
A. 52g of CCl4
B. Cl2
C. 4.44g
Step-by-step explanation:
Step 1:
The balanced equation for the reaction. This is given below:
CS2(s) + 4Cl2(g) —> CCl4(l) + 2SCl2(s)
Step 2:
Determination of the masses of CS2 and Cl2 that reacted and the mass of CCl4 produced from the balanced equation. This is illustrated below:
CS2(s) + 4Cl2(g) —> CCl4(l) + 2SCl2(s)
Molar Mass of CS2 = 12 + (32x2) = 76g/mol
Molar Mass of Cl2 = 2 x 35.5 = 71g/mol
Mass of Cl2 from the balanced equation = 4 x 71 = 284g
Molar Mass of CCl4 = 12 + (35.5x4) = 154g/mol
Summary:
From the balanced equation above,
76g of CS2 reacted with 284g of Cl2 to produce 154g of CCl4.
Step 3.
Determination of the limiting reactant.
From the balanced equation above,
76g of CS2 reacted with 284g of Cl2.
Therefore, 30.1g of CS2 will react with = (30.1 x 284)/76 = 112.48g of Cl2.
We can see that it requires a higher mass of Cl2 (112.48g) compare to 95.9g that was given to react with the 30.1g of CS2 given from the question. This simply means that Cl2 is the limiting reactant and CS2 is the excess reactant.
A. Determination of the maximum mass of CCl4 produced.
The limiting reactant is used to obtain the maximum amount.
From the balanced equation above,
284g of Cl2 to produce 154g of CCl4.
Therefore, 95.9g of Cl2 will produce = (95.9 x 154)/284 = 52g of CCl4.
Therefore, the maximum mass of CCl4 produced from the reaction is 52g
B. The limiting reagent is Cl2.
C. Determination of mass of the excess reagent that remains.
CS2 is the excess reagent. The mass of CS2 that remains can be obtained as follow:
From the balanced equation above,
76g of CS2 reacted with 284g of Cl2.
Therefore Xg of CS2 will react with 95.9g of Cl2 i.e
Xg of CS2 = (76 x 95.9)/284
Xg of CS2 = 25.66g.
Mass of CS2 given = 30.1g
Mass of CS2 that react = 25.66g
Mass of CS2 that remains = (Mass of CS2 given) - (Mass of CS2 that react)
Mass of CS2 that remains = 30.1 - 25.66
Mass of CS2 that remains = 4.44g