A+ 2B + 3C⇌ AB2C3 . Reaction of 6.0 g of atoms of B, and 0.036 mol of C yields 4.8 g of compound AB2C3 . If the atomic masses of A and C are 60 and 80 amu, respectively, the atomic mass B of is (Avogadro constant= 6 x 1023 mol-1)

A
70 amu
B
60 amu
C
50 amu
D
40 amu

Solution:

The given data are:

$\underset{6.0 g}{A} + \underset{6.0 x 10^{23} atoms}{2 B} + \underset{0.036 mol}{3 C} \equiv \underset{4.8 g}{{AB}_{2} C_{3}}$

$= \frac{6}{60} mol ≃ 1 mol$

Maximum extent of reaction $0.1 mol 0.5 mol 0.012 mol$

The limiting reagent is the atom $C$ . The amount of ${AB}_{2} C_{3}$ formed will also be 0.012 mol. Mass of ${AB}_{2} C_{3}$ formed is $4.8 g$ . Hence

$(0 .012 \times 60 + (0 .024 m o l) M_{R} + 036 \times 80) g = 4 .8 g$

This gives $M_{B} = 50 g {mol}^{- 1}$ Thus, atomic mass of B is $50 amu$ .

$A + 2 B + 3 C ⇌ {AB}_{2} C_{3}$ . Reaction of $6.0 g$ of atoms of $B$ , and $0.036 mol$ of $C$ yields $4.8 g$ of compound ${AB}_{2} C_{3}$ . If the atomic masses of $A$ and $C$ are $60$ and $80 amu$ , respectively, the atomic mass $B$ of is $(Avogadro constant = 6 x 10^{23} {mol}^{- 1})$

Solution:

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