Question

2.76 g of silver carbonate on being strongly heated yields a residue weighing:

[Ag = 108, C = 12, O = 16]

• A. 2.16 g
• B. 2.32 g
• C. 2.48 g
• D. 2.64 g

Answer 1

The correct option is A 2.16 g

Given,

$2.76g$ of ${Ag}_{2}C{O}_3$ was strongly heated, the reaction involved is:

$2{Ag}_{2}C{O}_3\left(s\right)⟶4Ag\left(s\right)+2C{O}_2\left(g\right)+O_2\left(g\right)$

As seen in the reaction, on heating ${Ag}_{2}C{O}_3$, $C{O}_2$ & $O_2$ are liberated, but $Ag$ is left behind as residue.

Now, Moles of $A{g}_{2}C{O}_3$ given=$\frac{2.76g}{276g/mol}=0.01$ moles.

$2$ moles $A{g}_{2}C{O}_3$ gives $4$ moles $Ag$

$0.01$ moles $A{g}_{2}C{O}_3$ gives $\frac{0.01\times 4}{2}$ moles $Ag$

$=0.02$ moles $Ag$

We know, $1$ mole $Ag=108g$

$\Rightarrow$ $0.02$ mole $Ag$

$=0.02\times 108=2.16gramAg$

Hence, option $A$ is correct.