Question

How many moles of ferric alum, $\left( NH _{4}\right)_{2} SO _{4} \cdot Fe _{2}\left( SO _{4}\right)_{3} \cdot 24 H _{2} O$ 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

Answer: (B)

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

$1 \,mol$ of alum $=2\, mol$ of $Fe$

$2 \,mol$ of $Fe =1 \,mol$ of alum

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

$=0.5 \times 10^{-4} mol$ of alum