How many moles of ferric alum, ( NH 4)2 SO 4 ⋅ Fe 2( SO 4)3 ⋅ 24 H 2 O can be made from the sample of Fe containing 0.0056 g of it?

Question

How many moles of ferric alum, ( NH 4)2 SO 4 ⋅ Fe 2( SO 4)3 ⋅ 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 × 10-4 mol (C) 0.33 × 10-4 mol (D) 2 × 10-4 mol

Tardigrade Admin · Accepted Answer

Moles of Fe =(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 =(1/2) × 10-4 mol of alum =0.5 × 10-4 mol of alum

Q. How many moles of ferric alum, $(N H_{4})_{2} S O_{4} \cdot F e_{2} (S O_{4})_{3} \cdot 24 H_{2} O$ can be made from the sample of $F e$ containing $0.0056 g$ of it?

$1 0^{- 4} m o l$

$0.5 \times 1 0^{- 4} m o l$

$0.33 \times 1 0^{- 4} m o l$

$2 \times 1 0^{- 4} m o l$

Moles of $F e = \frac{0.0056}{56} = 1 0^{- 4}$

$1 m o l$ of alum $= 2 m o l$ of $F e$

$2 m o l$ of $F e = 1 m o l$ of alum

$1 0^{- 4} m o l$ of $F e = \frac{1}{2} \times 1 0^{- 4} m o l$ of alum

$= 0.5 \times 1 0^{- 4} m o l$ of alum

Q. How many moles of ferric alum, (NH4​)2​SO4​⋅Fe2​(SO4​)3​⋅24H2​O can be made from the sample of Fe containing 0.0056g of it?

10−4mol

0.5×10−4mol

0.33×10−4mol

2×10−4mol