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(ix) Zn +HNO_3 to Zn (NO_3)_2 +NH_(4)NO_3+H_2O . |
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Answer» Solution :`Zn+HNO_3 to Zn(NO_3)_2 +NH_4NO_3+H_2O` `Zn^(0) to ZN^(2+)+2e^(-)` ……………(1) `N^(+5)+8e^(-) to N^(-3)` ……………(2) Multiply Equation (1) by 4 to balance the electorns `4Zn^(0) to 4Zn^(2+)+2e^(-)` ……………(3) ADD equation (3) and (2) `4Zn^(0)to 4Zn^(2+)+ cancel(8e^(1))` `N5+cancel(8e^1)to N^(3-)`. `4Zn^(0)+N^(5+)to 4Zn^(2+)+N^(2)` . Overall equation . `4Zn+10HNO_3 to 4Zn(NO_3)_2 +NH_4NO_3+H_2O` Balance all the ATOMS except O and H `4Zn +10HNO_(3)to 4Zn (NO_3)_2+ NH_4NO_3+H_2O` Balance oxygen atom by adding `H_2O` on the side FALLING short of oxygen atom . `4Zn+10HNO_3 to 4Zn(NO_3)_2 +NH_4NO_3+3H_2O` . |
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