Question

When $100ml$ of $0.06m$ Fe${\left({NO}_3\right)}_3,$ $50ml$ of $0.2ml$ $FeC{l}_3$ and $100ml$ of $0.26mmg{\left({NO}_3\right)}_2$ are mixed , then calculate molarity of reaction in final solution.

Answer 1

Total volume $=100+50+100=250ml$

$m$ moles of $F{e}^{3+}=100\times 0.06+50\times 0.2=16$

$\therefore \left[F{e}^{3+}\right]=\frac{16}{250}=0.064M$

$m$ moles of $N{O}_3^{-}=100\times 0.06\times 3+0.26\times 2\times 100=70$

$\therefore \left[N{O}_3^{-}\right]=\frac{70}{250}=0.28M$

$m$ moles of $C{l}^{-}=50\times 0.2\times 3=30$

$\therefore \left[C{l}^{-}\right]=\frac{30}{250}=0.12M$

$m$ moles of $M{g}^{2+}=100\times 0.26=26$

$\therefore \left[M{g}^{2+}\right]=\frac{26}{250}=0.104M$