The Claus reactions, 2H2S+3O2=2SO2+2H2O and SO2+2H2S=3S+2H2O, are used to generate elemental sulfur. How much sulfur can be produced from 48.0g of O2?

A good way to do such things is to couple the coefficients in the two reactions. In this case you need to make sure the coeff for SO2 is the same

The first reads

2 : 3 ==> 2SO2 : 2

Make the second one twice what it was

………… 2SO2 : 4 ===> 6S

Notice that 3O2 eventually results in 6S

So: 2SO2 ===> 6S

Or 2 :: 6

Now divide 48.0 g of O2 by the molar mass of O2 to get the number of mol of O2: 48/32 = 1.5 mol

Now divide the coeffs by 2:

1 :: 3

Then multiply by whatever you actually have (i.e. 1.5)

1.5 :: 4.5

So you’ll get 4.5 mole of S or 144 g if you prefer.

The best way to answer this question is to first look at the balanced chemical equations;-

2 H2S + 3 O2 → 2 SO2 + 2 H2O ……. (Equation #1)

SO2 + 2 H2S → 3S + 2 H2O ……. (Equation #2)

Now, the first equation produces 2 moles of SO2 while the second equation requires only one mole of SO2. So, to balance these consecutive equations with respect to SO2, we can divide the coefficients of the first equation by 2, which gives;--

H2S + 1.5 O2 → SO2 + H2O ……. (Equation #3)

Since the molar mass of O2 = 32, then 48g of O2 = 48/32 =1.5 moles of O2.

According to equation #3, 1.5 moles of O2 will produce 1 mole of SO2.

Since the molar mass of S = 32, then, according to equation #2, 1 mole of SO2 will produce 3 moles of S, or 3 * 32 = 96g.

So, the answer to the question is that, using the Claus equations given above, 48g of O2 will produce 96g of sulphur.