Question

For the following reaction, 13.6 grams of sulfur are allowed to react with 17.6 grams of carbon monoxide .
sulfur(s) + carbon monoxide(g) sulfur dioxide(g) + carbon(s)
What is the maximum amount of sulfur dioxide that can be formed?

What is the FORMULA for the limiting reagent?
What amount of the excess reagent remains after the reaction is complete?
grams

Answer 1

S(s) + 2CO(g) ------------> S0₂(g) + 2C(s)

1 mole S reacts with 2 moles CO giving 1 mole SO₂ and 2 moles C

In terms of weight,

32 g S reacts with 56 g CO giving 64 SO₂ and 24 g C.

$32/56<13.6/17.6\implies$ CO is the limiting reagent.

Using unitary method, 1g CO gives 1.143 g SO₂

17.6 g CO will give 20.114 g SO₂

Calculating left over sulphur,

1g SO₂ comes from 0.5 g S$\implies$ 20.114 g SO₂ come from 10.057 g S

$\therefore$ Leftover S = 3.543 g