Chemistry - Stoichiometry Question with Solution | TestHub

To achieve a yield greater than 50% with 25 g precipitate from 100 g reactant, the theoretical mass of precipitate from 100 g reactant must be less than 50 g.

(A) $2NaHC{O}_3\to N{a}_{2}C{O}_3\downarrow +C{O}_2\uparrow +H_{2}O\uparrow$

Molar mass of $2NaHC{O}_3=2\times (23+1+12+3\times 16)=168$ g/mol

Molar mass of $N{a}_{2}C{O}_3=2\times 23+12+3\times 16=106$ g/mol

From 168 g $NaHC{O}_3$, 106 g $N{a}_{2}C{O}_3$ is produced.

From 100 g $NaHC{O}_3$, theoretical $N{a}_{2}C{O}_3=\frac{106}{168}\times 100=63.1$ g.

Yield $=\frac{25}{63.1}\times 100=39.6\%$

(B) $NaN{O}_3\to NaN{O}_2\downarrow +\frac{1}{2}{O}_2\uparrow$

Molar mass of $NaN{O}_3=23+14+3\times 16=85$ g/mol

Molar mass of $NaN{O}_2=23+14+2\times 16=69$ g/mol

From 85 g $NaN{O}_3$, 69 g $NaN{O}_2$ is produced.

From 100 g $NaN{O}_3$, theoretical $NaN{O}_2=\frac{69}{85}\times 100=81.2$ g.

Yield $=\frac{25}{81.2}\times 100=30.8\%$

(C) $Ca(N{O}_3{)}_2\to CaO\downarrow +2N{O}_2\uparrow +\frac{1}{2}{O}_2\uparrow$

Molar mass of $Ca(N{O}_3{)}_2=40+2\times (14+3\times 16)=164$ g/mol

Molar mass of $CaO=40+16=56$ g/mol

From 164 g $Ca(N{O}_3{)}_2$, 56 g $CaO$ is produced.

From 100 g $Ca(N{O}_3{)}_2$, theoretical $CaO=\frac{56}{164}\times 100=34.1$ g.

Yield $=\frac{25}{34.1}\times 100=73.3\%$

(D) $PbC{O}_3\to PbO\downarrow +C{O}_2\uparrow$

Molar mass of $PbC{O}_3=208+12+3\times 16=268$ g/mol

Molar mass of $PbO=208+16=224$ g/mol

From 268 g $PbC{O}_3$, 224 g $PbO$ is produced.

From 100 g $PbC{O}_3$, theoretical $PbO=\frac{224}{268}\times 100=83.6$ g.

Yield $=\frac{25}{83.6}\times 100=29.9\%$

Only reaction (C) has a yield greater than 50%.

The final answer is $\boxed{C}$