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.

 

[DeeNos]

Hi there,

I can definitely help you with these questions! Let's break them down one by one.

1. To determine the limiting reactant, we need to compare the moles of each reactant present. Since we have the mass and purity of each reactant, we can calculate the moles of each as follows:

- Iron: 20.0 g x 0.60 purity = 12.0 g Fe
12.0 g Fe / 56.0 g/mol = 0.214 moles Fe
- Sulphuric acid: 20.0 g x 0.80 purity = 16.0 g H2SO4
16.0 g H2SO4 / 98.0 g/mol = 0.163 moles H2SO4

Since there are more moles of iron than sulphuric acid, iron is the limiting reactant.

2. To determine the amount of Iron (II) sulphate obtained, we need to use the mole ratio from the balanced equation. From the equation, we can see that for every 1 mole of Fe, we get 1 mole of FeSO4. Therefore, the amount of FeSO4 obtained will be equal to the amount of Fe used, which is 0.214 moles.

3. To determine the volume of H2 obtained, we need to use the ideal gas law: PV = nRT. We know the pressure (1 atm), temperature (r.t.p), and moles of H2 (0.214 moles). We also know the molar volume at r.t.p (24 dm^3/mol). Therefore, we can calculate the volume of H2 using the following equation:

V = (nRT)/P = (0.214 mol x 0.08206 L atm/mol K x 298 K)/1 atm = 5.27 L

4. If the yield were just 80%, we would need to multiply the moles of FeSO4 obtained by 0.80 to get the actual amount. So, the amount of FeSO4 obtained would be 0.214 moles x 0.80 = 0.171 moles. Then, we can use the ideal gas law again to calculate the volume of H2:

V = (nRT)/P = (0.171 mol x 0.08206 L atm/mol K x 298 K)/