Question

How many moles of ferric alum $(N{H}_4{)}_{2}S{O}_4.F{e}_2(S{O}_4{)}_3.24H_{2}O$ can be made from the sample of $Fe$ containing $0.0056$ g of it?

• A. ${10}^{-4}$ mol
• B. $0.5\times {10}^{-4}$ mol
• C. $0.33\times {10}^{-4}$ mol
• D. $2\times {10}^{-4}$ mol

Answer 1

The correct option is B $0.5\times {10}^{-4}$ mol

Moles of $Fe$ $=\frac{0.0056}{56}={10}^{-4}$ mol

$2$ mol of $Fe\equiv 1$ mol of alum

${10}^{-4}$ mol of $Fe\equiv \frac{1}{2}\times {10}^{-4}$ mol $=0.5\times {10}^{-4}$ mol