Question

Write the following equation in ionic form:
a) $Mn{O}_2+4HCl\to MnC{l}_2+2H_{2}O+C{l}_2$
b) $2NaHC{O}_3+H_{2}S{O}_4\to N{a}_{2}S{O}_4+H_{2}O+C{O}_2$

• A. a) $Mn{O}_2+4H^{+}+4C{l}^{-}\to M{n}^{2+}+2C{l}^{-}+2H_{2}O+C{l}_2$,
  b) $2N{a}^{+}+2{HCO}_3^{-}+2H^{+}+S{O}_4^{2-}\to 2N{a}^{+}+S{O}_4^{2-}+2H_{2}O+C{O}_2$
• B. a) $2Mn{O}_2+8H^{+}+4C{l}^{-}\to 2M{n}^{2+}+2C{l}^{-}+4H_{2}O+C{l}_2$,
  b) $2N{a}^{+}+2{HCO}_3^{-}+2H^{+}+S{O}_4^{2-}\to 2N{a}^{+}+S{O}_4^{2-}+2H_{2}O+C{O}_2$
• C. a) $Mn{O}_2+4H^{+}+3C{l}^{-}\to M{n}^{2+}+C{l}^{-}+2H_{2}O+C{l}_2$,
  b) $2N{a}^{+}+2{HCO}_3^{-}+2H^{+}+S{O}_4^{2-}\to 2N{a}^{+}+S{O}_4^{2-}+2H_{2}O+C{O}_2$
• D. None of these

Answer 1

Answer: (A)

a) $Mn{O}_2+4H^{+}+4C{l}^{-}\to M{n}^{2+}+2C{l}^{-}+2H_{2}O+C{l}_2$,

b) $2N{a}^{+}+2{HCO}_3^{-}+2H^{+}+S{O}_4^{2-}\to 2N{a}^{+}+S{O}_4^{2-}+2H_{2}O+C{O}_2$
(a) In the first reaction, $MnC{l}_2$ and $HCl$ are ionic in nature.
Therefore, the reaction in ionic form is as follows:
$\therefore Mn{O}_2+4H^{+}+4C{l}^{-}\to M{n}^{2+}+2C{l}^{-}+2H_{2}O+C{l}_2$ (Taking $2C{l}^{-}$ common)
(b) In the second reaction, $NaHC{O}_3,H_{2}S{O}_4$ and $N{a}_{2}S{O}_4$ are ionic in nature.
Therefore, the reaction in ionic form is as follows:
$2N{a}^{+}+2{HCO}_3^{-}+2H^{+}+S{O}_4^{2-}\to 2N{a}^{+}+S{O}_4^{2-}+2H_{2}O+C{O}_2$ (Taking $S{O}_4^{2-}$ common) or $2N{a}^{+}+2{HCO}_3^{-}+2H^{+}+S{O}_4^{2-}\to 2N{a}^{+}+2H_{2}O+C{O}_2$.