Can be recovered 496.7 kilograms of iron from 639 kilograms of the ore FeO.
To find the mass of Fe in the ore FeO, we need to calculate the number of moles of the ore.
[tex] n_{FeO} = \frac{m_{FeO}}{M_{FeO}} [/tex]
Where:
[tex] m_{FeO}[/tex]: is the mass of FeO = 639 kg = 639000 g
[tex] M_{FeO}[/tex]: is the molar mass of FeO = 71.844 g/mol
The number of moles of FeO is:
[tex] n_{FeO} = \frac{m_{FeO}}{M_{FeO}} = \frac{639000 g}{71.844 g/mol} = 8894.3 \:moles [/tex]
Now, in 1 mol of FeO we have 1 mol of Fe, so the number of moles of Fe is:
[tex] n_{Fe} = n_{FeO} = 8894.3 \:moles [/tex]
Then, the mass of Fe is:
[tex] m_{Fe} = n_{Fe}*A_{Fe} = 8894.3 \:moles*55.84 g/mol = 496657.7 g = 496.7 kg [/tex]
Therefore, 496.7 kilograms of iron can be recovered.
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I hope it helps you!
