You visited us 6 times! Enjoying our articles? Unlock Full Access!

Stoichiometric Calculations

A metal M f...

Question

A metal $M$ forms the sulphate $M_{2} (S O_{4})_{3}$ . A $0.596 g$ sample of the sulphate reacts with excess $B a C l_{2}$ to give $1.220 g B a S O_{4}$ . What is the atomic weight of $M$ ? (Atomic weights: $S = 32, B a = 137.3$ )

26.9

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

69.7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

55.8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

23

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $26.9$
Moles of $B a S O_{4} = \frac{1.22}{233.3}$
Moles of $M_{2} (S O_{4})_{3} = \frac{1.22}{233.3} \times \frac{1}{3}$ $= 1.743 \times 10^{- 3}$
Weight of $M_{2} (S O_{4})_{3} = 0.596$
$∴ 1.743 \times 10^{- 3} (2 M + 96 \times 3) = 0.596$
Hence, $M = 26.9$

Suggest Corrections

Similar questions

M

M g S O_{4} . 7 H_{2} O

M

M

B a C l_{2}

24.4 g

46.6 g

23.4 g

B a C l_{2} + N a_{2} S O_{4} \to B a S O_{4} + 2 N a C l

B a C l_{2}

24.4 g

46.6 g

23.4 g

B a C l_{2} + N a_{2} S O_{4} \to B a S O_{4} + 2 N a C l

N a_{2} S O_{4} + B a C l_{2} ⟶ B a S O_{4} + 2 N a C l

[N a = 23, O = 16, S = 32, B a = 132]

Join BYJU'S Learning Program