Question

Find equivalent weight of underline elements in the following compounds:
(i) ${\underset{–}{Fe}}_2{O}_3$ (ii) $\underset{–}{Ca}O$
(iii) $H\underset{–}{Br}$ (iv) $\underset{–}{Fe}O$
(v) ${\underset{–}{Ag}}_{2}S$ (vi) ${\underset{–}{Al}}_2{O}_3$
(vii) $\underset{–}{Mg}C{l}_2$ (viii) $\underset{–}{Cu}O$
(ix) ${\underset{–}{Cu}}_{2}C{l}_2$ (x) $\underset{–}{Pb}O$
(xi) $\underset{–}{Pb}{O}_2$ (xii) ${\underset{–}{Pb}}_3{O}_4$
(xiii) $F{e}_3{O}_4$ (xiv) ${\underset{–}{PH}}_3$
(xv) ${\underset{–}{N}}_{3}H$ (xvi) ${\underset{–}{Fe}}_{0.93}O$
(xvii)${\underset{–}{Sn}}_{4.44}{O}_8$ (xviii) $\underset{–}{Ca}{H}_2$
(xix) ${\underset{–}{P}}_2{H}_4$ (xx) $\underset{–}{Sn}{O}_2$

Answer 1

(i) ${\underset{–}{Fe}}_2{O}_3\to oxi.state=+3,Eq.Wt.=\frac{M}{3}$

(iii) $H\underset{–}{Br}\to oxi.state=-1,Eq.Wt.=\frac{M}{1}$

(v) ${\underset{–}{Ag}}_{2}S\to oxi.state=+1,Eq.Wt.=\frac{M}{1}$

(vii) $\underset{–}{Mg}C{l}_2\to oxi.state=+2,Eq.Wt.=\frac{M}{2}$

(ix) ${\underset{–}{Cu}}_{2}C{l}_2\to oxi.state=+1,Eq.Wt.=\frac{M}{1}$

(xi) $\underset{–}{Pb}{O}_2\to oxi.state=+4,Eq.Wt.=\frac{M}{4}$

(xiii) $F{e}_3{O}_4\to FeO+F{e}_2{O}_3$, Oxi. state +2 and +3$

$Eq.Wt.=\frac{M}{2}+\frac{M}{3}$

(xv) ${\underset{–}{N}}_{3}H\to oxi.state=\frac{-1}{3},Eq.Wt.=3M$

(xvii)${\underset{–}{Sn}}_{4.44}{O}_8\to oxi.state=+4,Eq.Wt.=\frac{M}{4}$

(xix) ${\underset{–}{P}}_2{H}_4\to oxi.state=-4,Eq.Wt.=\frac{M}{4}$

(iv) $\underset{–}{Fe}O\to Oxi.state==+2,Eq.Wt.=\frac{M}{2}$

(vi) ${\underset{–}{Al}}_2{O}_3\to Oxi.state=+3,Eq.Wt.=\frac{M}{3}$

(viii) $\underset{–}{Cu}O\to Oxi.state=+2,Eq.Wt.=\frac{M}{2}$

(x) $\underset{–}{Pb}O\to Oxi.state=+2,Eq.Wt.=\frac{M}{2}$

(xii) ${\underset{–}{Pb}}_3{O}_4\to Oxi.state=+\frac{8}{3},Eq.Wt.=\frac{3M}{8}$

(xiv) ${\underset{–}{PH}}_3\to Oxi.state=-3Eq.Wt.=\frac{M}{3}$

(xvi) ${\underset{–}{Fe}}_{0.93}O\to oxi.state=+2,Eq.Wt.=\frac{M}{2}$

$InF{e}_{.93}O$, some ions of $F{e}^{+2}$ are disappeared.

(xviii) $\underset{–}{Ca}{H}_2\to Oxi.state=+2,Eq.Wt.=\frac{M}{2}$

(xx) $\underset{–}{Sn}{O}_2\to Oxi.state=+4,Eq.Wt.=\frac{M}{4}$