What amount of silver will be obtained on thermal decomposition of 2.76 g of silver carbonate, $A g_{2} C O_{3}$ (276 amu) ?

2.16 g

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0.216 g

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216 g

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21.6 g

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Solution

The correct option is A 2.16 g
On thermal decomposition, we have
$A g_{2} C O_{3} \to 2 A g + C O_{2} + 1 / 2 O_{2}$
1 mole of $A g_{2} C O_{3}$ produce 2 mole of Ag
moles of $A g_{2} C O_{3}$ = $\frac{g i v e n w e i g h t}{m o l e c u l a r w e i g h t}$ = $\frac{2.76}{276}$ = 0.01
so, moles of Ag produced = $2 * 0.01$ = 0.02
amount of Ag = $0.02 * a t o m i c m a s s$ = $0.02 * 108$ = 2.16 g

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Similar questions

2.76 g

A g = 108 g/mol

A g_{2} C O_{3} Δ - \to 2 A g + C O_{2} + \frac{1}{2} O_{2}

2.76g of silver carbonate (at mass of Ag = 108) on being heated strongly yields a residue

2.76 g

A g = 108 g/mol

A g_{2} C O_{3} Δ - \to 2 A g + C O_{2} + \frac{1}{2} O_{2}

2.76 g

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