Answer: The mass of iron in the ore is 10.9 g
Explanation:
We are given:
Mass of iron (III) oxide = 15.6 g
We know that:
Molar mass of Iron (III) oxide = 159.69 g/mol
Molar mass of iron atom = 55.85 g/mol
As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.
To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:
In 159.69 g of iron (III) oxide, mass of iron present is [tex](2\times 55.85)=111.7g[/tex]
So, in 15.6 g of iron (III) oxide, mass of iron present will be = [tex]\frac{111.7g}{159.69g}\times 15.6g=10.9g[/tex]
Hence, the mass of iron in the ore is 10.9 g