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Question

Write the following equation in ionic form:
a) $$MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2$$
b) $$2NaHCO_3 + H_2SO_4 \rightarrow  Na_2SO_4 + H_2O+CO_2$$


A
a) MnO2+4H++4ClMn2++2Cl+2H2O+Cl2,
b) 2Na++2HCO3+2H++SO242Na++SO24+2H2O+CO2
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B
a) 2MnO2+8H++4Cl2Mn2++2Cl+4H2O+Cl2,
 
b) 2Na++2HCO3+2H++SO242Na++SO24+2H2O+CO2
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C
a) MnO2+4H++3ClMn2++Cl+2H2O+Cl2,
b) 2Na++2HCO3+2H++SO242Na++SO24+2H2O+CO2
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D
None of these
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Solution

The correct option is C a) $$MnO_2 + 4H^++ 4Cl^- \rightarrow Mn^{2+} + 2Cl^- + 2H_2O + Cl_2$$,

b) $$2Na^+ + 2{HCO}^{-}_3 + 2H^+ + SO^{2-}_4 \rightarrow 2Na^+ + SO^{2-}_4 + 2H_2O + CO_2$$
(a) In the first reaction, $$MnCl_2$$ and $$HCl$$ are ionic in nature.
Therefore, the reaction in ionic form is as follows:
$$ \therefore  MnO_2 + 4H^++ 4Cl^- \rightarrow Mn^{2+} + 2Cl^- + 2H_2O + Cl_2$$ (Taking $$2Cl^-$$ common)
(b) In the second reaction, $$NaHCO_3, H_2SO_4$$ and $$Na_2SO_4$$ are ionic in nature.
Therefore, the reaction in ionic form is as follows:
$$2Na^+ + 2{HCO}^{-}_3 + 2H^+ + SO^{2-}_4 \rightarrow 2Na^+ + SO^{2-}_4 + 2H_2O + CO_2$$ (Taking $$SO^{2-}_4$$ common) or $$2Na^+ + 2{HCO}^{-}_3 + 2H^+ + SO^{2-}_4 \rightarrow 2Na^+ + 2H_2O + CO_2$$.

Chemistry

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