how many grams of iron will be required to release all the antimony from 10 grams of antimony trisulfide? iron 2 sulfide is also formed

mass of Fe = 4.86 g

Explanation:

Balanced chemical equation

Sb₂S₃ + 3Fe → 2Sb + 3FeS

Given data from eq.

moles of iron = 3 mol

mas of iron = ?

moles of antimony trisulfide = 1 mol

mass of antimony trisulfide = 10 g

Solution

1st we will find out the mole ratio of Sb₂S₃ and Fe from balanced chemical equation

Sb₂S₃ : Fe

1 : 3

Now we find out number of moles of citric acid Sb₂S₃ from the given mass

moles = mass / molar mass

moles = 10 g / 339.7 g/mol

moles = 0.029 mol

Now we find out moles (x) of iron needed for 0.029 mol of antimony trisulfide

from the balanced chemical equation the mole ratios are:

1 : 3

0.029 : x

Cross multiply these ratios

3 * 0.029 = 1x

0.087 = 1 x

x = 0.087 mol

Now we will find out the mass of iron

mass = moles * molar mass

mass of Fe = 0.087 mol * 55.845 g/mol

mass of Fe = 4.86 g