Chemistry doubt

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1.26K viewsApril 18, 2024

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Dhanmasih Bage April 18, 2024 0 Comments

Calculate the mass of iron in \(10~Kg\) of iron ore which contains \(80\%\) of pure ferric oxide.

score 1 · Answer 1 · 2024-04-18T09:15:15+00:00

Solution:-

Amount of \(\ce{Fe2O3}\) present in the ore = \(\frac{\normalsize80}{\normalsize100}\times~10~Kg\)

= \(8~Kg\)

Molecular Mass of \(\ce{Fe2O3}\) = \(2\times56 + 3\times16\)

= \(112 +48\)

= \(160~amu\)

\(\%\) of \(\ce{Fe}\) in \(\ce{Fe2O3}\)= \(\frac{\normalsize112}{\normalsize160}\)\(\times100\)

= \(70\%\)

\(\therefore\) \(\ce{Fe}\) present in \(\ce{Fe2O3}\) = \(70\%\) of \( 8~Kg\)

= \(\frac{\normalsize70}{\normalsize100}\times 8000~g\)

= \(5600~g\)

= \(56~Kg\)