Respuesta :
Molar mass of iron(III)sulfate that is Fe₂(SO₄)₃ =399.88 g mol⁻¹
Mass given of iron(III)sulfate that is Fe₂(SO₄)₃= 0.275 g
Number of moles of iron(III)sulfate that is Fe₂(SO₄)₃= [tex]\frac{0.275}{399.88}[/tex]
=0.00068 mol
there are two moles of Fe in 1 mole of Fe₂(SO₄)₃
So, in 0.00068 mol of Fe₂(SO₄)₃=2 ×0.00068 mol of Fe
= 0.00136 mol of Fe
molar mass of Fe=55.8 g mol⁻¹
So the mass of iron in 0.00136 mol of Fe= 55.8 g mol⁻¹ × 0.00136 mol
=0.075 g
Mass of iron = [tex]0.275 grams[/tex]
The chemical formula of iron (III) sulfate = [tex]Fe_{2}(SO_{4})_{3}[/tex]
First, calculate the number of moles of [tex]Fe_{2}(SO_{4})_{3}[/tex]
Number of moles of iron (III) sulfate = [tex]\frac{given mass in g}{molar mass}[/tex]
Molar mass of iron (III) sulfate = 399.88 g/mol
Number of moles of iron (III) sulfate = [tex]\frac{0.275 g}{399.88 g/mol}[/tex]
= [tex]6.87\times 10^{-4} mol[/tex]
In one mole of iron (III) sulfate there are two moles of iron according to the chemical formula.
So, in [tex]6.87\times 10^{-4} mol[/tex] of iron (III) sulfate there are =[tex] 2\times 6.87 \times 10^{-4} mol[/tex] of iron.
Number of moles of iron =[tex]13.74\times 10^{-4} mol[/tex]
Molar mass of iron = [tex]55.8 g/mol[/tex]
Mass of iron =[tex]number of moles \times molar mass[/tex]
= [tex]13.74\times 10^{-4} mol \times 55.8 g/mol[/tex]
= [tex]766.692 \times 10^{-4} g[/tex]
= [tex]7.66 \times 10^{-2} g[/tex]
Hence, mass of iron = [tex]7.66 \times 10^{-2} g[/tex]
