MCQ
How many moles of ferric-alum $(NH_4)_2SO_4.Fe_2(SO_4)_3.24H_2O $ can be made from the sample of $Fe$ containing $0.0056\,g$ of it ?
- A$10^{-4}\,mol$
- ✓$0.5\times 10^{-4}\,mol$
- C$0.33\times 10^{-4}\,mol$
- D$2\times 10^{-4}\,mol$
✓
Answer
Correct option: B.
$0.5\times 10^{-4}\,mol$
b
Moles of $\mathrm{Fe}=\frac{0.0056}{56}=10^{-4} \,\mathrm{mol}$
Moles of $\mathrm{Fe}=\frac{0.0056}{56}=10^{-4} \,\mathrm{mol}$
$1\, \mathrm{mol}$ of $\mathrm{alum}=2\, \mathrm{mol}$ of $\mathrm{Fe}$
$2\, \mathrm{mol}$ of $\mathrm{Fe}=1\, \mathrm{mol}$ of alum
$10^{-4} \,\mathrm{mol}$ of $\mathrm{Fe}=\frac{1}{2} \times 10^{-4}\, \mathrm{mol}$
$=0.5 \times 10^{-4}\, \mathrm{mol}$
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