Question

If 7.0 mol hydrogen sulfide are reacted with 3.0 mol sulfur dioxide, how many moles of sulfur (S) can be formed (the other product is water)? How many moles of which reagent are left?

Answer 1

1)

Balanced chemical equation is:

H2S + 2 SO2 ---> 3 S + 2 H2O

1 mol of H2S reacts with 2 mol of SO2

for 7 mol of H2S, 14 mol of SO2 is required

But we have 3 mol of SO2

so, SO2 is limiting reagent

we will use SO2 in further calculation

According to balanced equation

mol of S formed = (3/2)* moles of SO2

= (3/2)*3

= 4.5 mol

Answer: 4.5 mol of S will be formed

2)

According to balanced equation

mol of H2S reacted = (1/2)* moles of SO2

= (1/2)*3

= 1.5 mol

mol of H2S remaining = mol initially present - mol reacted

mol of H2S remaining = 7 - 1.5

mol of H2S remaining = 5.5 mol

Answer: 5.5 mol of H2S remains