  A mixture of 56.0 g of S and 1.10×102 g of Cl2 reacts completely to form S2Cl2 and SCl2.  Part AFind the mass of S2Cl2 formed

Question
A mixture of 56.0 g of S and 1.10×102 g of Cl2 reacts completely to form S2Cl2 and SCl2.

Part A

Find the mass of S2Cl2 formed
Step 1

Given:

Mass of S = 56.0 g

Mass of Cl2 = 1.10×102 g

We need to find the mass of S2Cl2

Step 2

Write the chemical reaction:

3S + 2Cl2 ---> S2Cl2 +SCl2

As per the chemical reaction,

3 mol. Of S (96g) reacts with 2 mol. Of Cl2 ( 142 g) of chlorine

The molar ratio is 3 : 2

And in the given question,

56 g of S reacts with 110 g of Cl2

32 g is 1 mol.

56g ---------? Mol.

56 X 1 / 32 = 1.75 mol.

1 mol. Of Cl= 71 g

? mol. = 110 g of Cl2

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.

Step 3

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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