Hi guys, im having some doubts getting this exercise:20,0 g of

[ignacioserra]

Hi guys, I am having some doubts getting this exercise:

20,0 g of iron (60% purity) are reacted with sulphuric acid (80 % pure) Iron (II) sulphate and hydrogen are formed:

Fe + H2SO4 → H2 + FeSO4

1. Which is the limiting reactant?
2. How much Iron (II) sulphate will be obtained?
3. What volume (r.t.p) of H2 (g) will be obtained?
4. What volume would be obtained if the yield were just 80 %?
5. How much iron should be used in this case if the amount calculated in (2) were to be obtained=

Info
Ar(H)= 1 Ar(O)= 16 Ar (Fe)= 56
Molar volume (r.t.p) 24dm^3

So,
Iron(II) Sulphate mass= 56 + 32 + (16x4) = 152
is that the iron sulphate obtained?

answer to number 5 should the be: 152/0.8?

please, I am needing some help!

 

[Borek]

How much iron in the sample? How many moles is it?

But there is something wrong - you are not told how much sulfuric acid was present, so question about a limiting reagent doesn't make sense.

 

[ignacioserra]

20 g of iron.
should I do 152/20?

 

[Borek]

20 g of iron.

No, there is no 20 g of iron. There is a 20g sample that contains 60% iron, that's not the same.

should I do 152/20?

Guessing won't get you far.

You have to follow the stoichiometry. That means calculating mass of the iron, calculating number of moles of the iron, calculating number of moles of products (from the reaction stoichiometry), then converting these moles to mass and volume. In this exact order.