if sulfur is in excess then production of 2 moles of ferrous sulfide requires how many gram of iron?

Ferrous sulfide is FeS, so each mole of ferrous sulfide requires 1 mole of iron; hence the answer is 2 gram-atoms of iron, i.e., about 112 grams of iron.

Well, well, well, looks like we've got a stoichiometry problem on our hands! To find out how many grams of iron are needed to produce 2 moles of ferrous sulfide, we need to know the molar ratio between iron and ferrous sulfide.

Let's take a peek at the balanced chemical equation for the reaction:
Fe + S → FeS

According to this equation, we can see that 1 mole of iron reacts with 1 mole of sulfur to produce 1 mole of ferrous sulfide.

So, if we want to produce 2 moles of ferrous sulfide, we would need 2 moles of iron.

But wait, there's more! We need to convert moles to grams. The molar mass of iron (Fe) is approximately 55.845 g/mol.

So, to calculate the grams of iron required, we multiply the number of moles by the molar mass:
2 moles Fe * 55.845 g/mol = 111.69 grams of iron

So, if we have an excess of sulfur, we would need approximately 111.69 grams of iron to produce 2 moles of ferrous sulfide.

To determine the number of grams of iron required to produce 2 moles of ferrous sulfide, we need to know the molar ratio between iron and ferrous sulfide.

The chemical formula for ferrous sulfide is FeS, which means it contains one iron atom (Fe) and one sulfur atom (S).

Using the molar masses of iron (Fe) and sulfur (S):

- Molar mass of Fe: 55.845 g/mol
- Molar mass of S: 32.06 g/mol

We can calculate the molar mass of ferrous sulfide (FeS):

Molar mass of FeS = (Molar mass of Fe) + (Molar mass of S)
= 55.845 g/mol + 32.06 g/mol
= 87.905 g/mol

Now, we can determine the molar ratio between iron (Fe) and ferrous sulfide (FeS) by comparing their coefficients in the balanced chemical equation:

Fe + S → FeS

From the equation, we can see that one mole of iron reacts with one mole of sulfur to produce one mole of ferrous sulfide.

Therefore, the molar ratio of iron to ferrous sulfide is 1:1.

Since the molar ratio is 1:1, we need the same number of moles of iron as moles of ferrous sulfide.

Given that we want to produce 2 moles of ferrous sulfide, we need 2 moles of iron.

Now, we can calculate the mass of iron required:

Mass of iron = (Number of moles of iron) x (Molar mass of iron)
= 2 moles x 55.845 g/mol
= 111.69 g

Therefore, if sulfur is in excess, the production of 2 moles of ferrous sulfide requires 111.69 grams of iron.

To determine the number of grams of iron required to produce 2 moles of ferrous sulfide when sulfur is in excess, we need to calculate the molar mass of ferrous sulfide and use stoichiometry.

1. Molar mass of FeS:
- The molar mass of iron (Fe) is 55.845 g/mol.
- The molar mass of sulfur (S) is 32.06 g/mol.
- Therefore, the molar mass of FeS = (55.845 g/mol) + (32.06 g/mol) = 87.905 g/mol.

2. Stoichiometry:
- The balanced equation for the reaction between iron and sulfur to form ferrous sulfide is:
Fe + S -> FeS
- The stoichiometric ratio between Fe and FeS is 1:1. This means that 1 mole of Fe reacts with 1 mole of FeS.
- Therefore, if 2 moles of FeS are produced, we need an equal number of moles of Fe.

3. Calculation:
- Moles of Fe = Moles of FeS = 2 moles.
- From the stoichiometric ratio, we know that 2 moles of FeS require 2 moles of Fe.
- To convert moles of Fe to grams, we need to multiply by the molar mass of Fe:
Grams of Fe = Moles of Fe × Molar mass of Fe
= 2 moles × 55.845 g/mol = 111.69 grams of Fe.

Therefore, if sulfur is in excess, the production of 2 moles of ferrous sulfide requires 111.69 grams of iron.