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Numerade Educator



Problem 27 Hard Difficulty

Iron reacts slowly with oxygen and water to form a compound commonly called rust $\left(\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 4 \mathrm{H}_{2} \mathrm{O}\right) .$ For 45.2 $\mathrm{kg}$ of rust, calculate (a) moles of compound; (b) moles of $\mathrm{Fe}_{2} \mathrm{O}_{3} ;(\mathrm{c})$ grams of $\mathrm{Fe} .$


$\mathrm{n}\left(\mathrm{Fe}_{2} \mathrm{O}_{3}^{*} 4 \mathrm{H}_{2} \mathrm{O}\right)=195,1 \mathrm{mol}$
$\mathrm{n}\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)=\mathrm{n}\left(\mathrm{Fe}_{2} \mathrm{O}_{3}{ }^{*} 4 \mathrm{H}_{2} \mathrm{O}\right)=195,1 \mathrm{mol}$
$\mathrm{m}(\mathrm{Fe})=21788,398 \mathrm{g}$


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