In acid medium Zn reduces nitrate ion to `NH_4^(+)` ion according to the reaction `Zn+NO_3^(-) to Zn^(2+)+ NH_4^(+) +H_2O` How many moles of HCL are required to reduce half a mole of `NaNO_3` completely? Assume the availability of sufficient Zn

Options

A. a) 5
B. b) 4
C. c) 3
D. d) 2

Answer

A

Explanation

First the given unbalanced equation is balanced by using following steps
Step I. The equation is splitted into two half equation as
`Zn to Zn^(2+), NO_3^(-) to NH_4^(+)`
Step II. Now water molecules are added to the side deficient in oxygen and `H^+` are added to the side deficient in hydrogen gas
`ZN to ZN^(2+), NO_3^(-) +10H^+ to NH_4^(+) +3H_2O`
Step III. THe number of electrons are balanced and the two half equations are added.
`[Zn to Zn^(2+)+2e^-] times 4, NO_3^(-)+10H^+ +8e^(-) to NH_4^(+) +3H_2O `
`4Zn to 4Zn^(2+) +8e^(-)`
`therefore 4Zn+NO_3^(-) +10H^+ to 4Zn^(2+)+ NH_4^(+)+3H_2O` (Net equation)
or `4Zn+NO_3^(-) +10HCl to 4Z^(2+) +NH_4^(+) +5Cl_2+3H_2O`
`because` 1 mole of `NO_3^(-)` is reduced by =10 moles of HCl
`therefore 1/2` mole of `NO_3^(-)` will be reduced by
`=10 times 1/2` moles of Hcl
=5 moles of HCl