Question

Mass of $A{g}_{2}C{O}_3$ is formed when $25.0$ mL of $0.2$ M $AgN{O}_3$ is mixed with $50.0$ mL of $0.08$ M $N{a}_{2}C{O}_3$ is written as $y\times {10}^{-2}$ g, then $y$ is :

Answer 1

The balanced chemical reaction is $2AgN{O}_3\left(2mol\right)+N{a}_{2}C{O}_3\left(1mol\right)\to A{g}_{2}C{O}_3\left(1mol\right)+2NaN{O}_3$.

The volume of silver nitrate is $25.0$ mL or $0.025$ L. Its molarity is $0.2$ M.

The number of moles of silver nitrate present $=0.025\times 0.2=0.005$ moles.

The volume of sodium carbonate is $50.0$ mL or $0.050$ L. Its molarity is $0.08$ M.

The number of moles of sodium carbonate $=0.050\times 0.08=0.004$ moles. They will require $0.008$ moles of silver nitrate.

However, only $0.005$ moles of silver nitrate are present.

Hence, silver nitrate is the limiting reagent.

The number of moles of silver carbonate formed is $0.5\times 0.005=0.0025$ moles.

The molar mass of silver carbonate is $275.7$ g/mol.

This corresponds to $0.0025\times 275.7=0.69$ g $=69\times {10}^{-2}$ g or $y\times {10}^{-2}$ g.

Hence, $y=69$.