When $3 g$ of a certain metal sulphide was roasted in air, $2.5 g$ of metal oxide was formed and along with $S O_{2}$ gas release. If oxidation number of metal is $+ 2$ . Calculate molar mass of metal.

64.00

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64.0

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Solution

Let suppose metal sulphide = $M S$

Metal oxide = $M O$

$+ 2 - 2 M S +_{2} \to + 2 - 2 M O + + 4 S O_{2}$

Equivalent of metal sulphide = Equivalent of metal oxide

$\frac{Mass of MS}{Molar mass of MS} \times n_{f} of M S = \frac{Mass of MO}{Molar moass of MO} \times n_{f} of M O$

$\Rightarrow \frac{3}{M + 32} = \frac{2.5}{M + 16}$

$\Rightarrow 3 M + 48 = 2.5 M + 80$

$\Rightarrow 3 M - 2.5 M = 80 - 48$

$\Rightarrow M = \frac{32}{0.5} = 64$

Therefore molar mass of metal is $64$ .

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When 3 g of a certain metal sulphide was roasted in air, 2.5 g of metal oxide was formed and along with SO2 gas release. If oxidation number of metal is +2. Calculate molar mass of metal.

When $3 g$ of a certain metal sulphide was roasted in air, $2.5 g$ of metal oxide was formed and along with $S O_{2}$ gas release. If oxidation number of metal is $+ 2$ . Calculate molar mass of metal.