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
Comments
Leave a comment