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How many moles of ferric alum${\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{.F}}{{\text{e}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}}{\text{.24}}{{\text{H}}_{\text{2}}}{\text{O}}$can be made from the sample of Fe containing $0.0056\,{\text{g}}$of it?
A. ${10^{ - 4}}\,{\text{mol}}$
B.$0.5\, \times \,{10^{ - 4}}\,{\text{mol}}$
C.$0.33\, \times \,{10^{ - 4}}\,{\text{mol}}$
D.$2\, \times \,{10^{ - 4}}\,{\text{mol}}$

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Last updated date: 20th Jun 2024
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Answer
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Hint: To determine the number of moles of an alum number of moles of iron is required. We can determine the mole of iron by using the mole formula. Mole formula gives the relation in mass, molar mass, and moles of a substance. After determining the moles of iron, by comparing the number of moles of iron and alum, the moles of alum can be determined.

Complete Step by step answer:We will use the mole formula to determine the mole of Fe, present in $0.0056\,{\text{g}}$.
The mole formula is as follows:
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
The molar mass of the iron is$56\,{\text{g/mol}}$.
Substitute $56\,{\text{g/mol}}$for molar mass and$0.0056\,{\text{g}}$for mass of iron.
${\text{mole}}\,\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.0056}}\,{\text{g}}}}{{56\,{\text{g/mol}}}}$
${\text{mole}}\,\,{\text{ = }}\,{10^{ - 4}}$
So, $0.0056$ grams of iron have ${10^{ - 4}}$ mole iron.
According to the formula of alum, we can say for the formation of one mole of alum two moles of Fe atoms are required. Two moles of iron give one, mole alum so, ${10^{ - 4}}$ mole of iron will give,
$2\,{\text{mol}}\,\,{\text{Fe}}\,{\text{ = }}\,1\,{\text{mol}}\,\,{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{.F}}{{\text{e}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}}{\text{.24}}{{\text{H}}_{\text{2}}}{\text{O}}$
${10^{ - 4}}\,{\text{mol}}\,\,{\text{Fe}}\,{\text{ = }}\,\,0.5\, \times \,{10^{ - 4}}\,{\text{mol}}\,{\left( {{\text{N}}{{\text{H}}_{\text{4}}}} \right)_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{.F}}{{\text{e}}_{\text{2}}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{3}}}{\text{.24}}{{\text{H}}_{\text{2}}}{\text{O}}$
So, ${10^{ - 4}}$ mole of iron will give, $\,0.5\, \times \,{10^{ - 4}}$mole alum.

Therefore, option (B) $\,0.5\, \times \,{10^{ - 4}}$ mole is correct.

Note: After the calculation of moles the number of atoms can be determined. For this multiply the moles with the Avogadro number. According to Avogadro, one mole of any substance contains $\,6.023\, \times \,{10^{23}}$ atoms, molecules, ions, or particles. Molecular mass for an element is the sum of the number of protons and neutrons. The molecular mass of the compound is the sum of the mass of all atoms present in the compound.