A mixture of 56.0 g of S and 1.10×10^{2} g of Cl2 reacts completely to form S2Cl2 and SCl2.

Part AFind the mass of S2Cl2 formed

Given:
Mass of S = 56.0 g
Mass of Cl_{2} = 1.10×10^{2} g
We need to find the mass of S_{2}Cl_{2}
Write the chemical reaction:
3S + 2Cl_{2} > S_{2}Cl_{2 }+SCl_{2}
As per the chemical reaction,
3 mol. Of S (96g) reacts with 2 mol. Of Cl_{2} ( 142 g) of chlorine
The molar ratio is 3 : 2
And in the given question,
56 g of S reacts with 110 g of Cl_{2 }
32 g is 1 mol.
56g ? Mol.
56 X 1 / 32 = 1.75 mol.
1 mol. Of Cl_{2 }= 71 g
? mol. = 110 g of Cl_{2}
110 X 1 / 71 = 1.54 mol.
Molar ratio of S: Cl = 1.75 : 1.54
From the above, it’s clear that S is limiting reagent.
Mass of S2Cl2 :
56 g S (1 mol S/32 g S) ( 1 mol S2Cl2/3mol S) = 0.577 mol.
110 g Cl2 (1 mol Cl2/70.90 g Cl2) ( 1 molS2Cl2/2 mol Cl2) = 0.77 mol.
mass of S2Cl2 formed = (0.577 mol.) ( 135 g S2Cl2 /1 mol S2Cl2)
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