Single Choice

$$\ln$$ the reduction of $$HNO_{3}$$ to $$N_{2}O$$ the no.of moles of electrons involved per mole of $$HNO_{3}$$ is :

A8
B4
Correct Answer
C3
D6

Solution

$$HNO_3 \rightarrow 0.5N_2O$$
Oxidation number of nitrogen in $$HNO_3$$ and $$N_2O$$ is +5 and +1 respectively so number of moles of electrons per mole of $$HNO_3$$ is 4.

SIMILAR QUESTIONS

p - block Elements

The formation of nitroglycerin is done by the use of concentrated nitric acid and concentrated sulphuric acid. The process of conversion of glycerine into nitro glycerine is termed as:

p - block Elements

Choose the strongest oxidizing pair from the following :

p - block Elements

The reaction of zinc with dilute and concentrated nitric acid, respectively, produces:

p - block Elements

How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M $$HNO_{3}$$? The concentrated acid is 70% $$HNO_{3}$$.

p - block Elements

With reference to aqura regia, choose the correct options(s)

p - block Elements

Among the metals $$Fe; Zn; Pb; Ag$$ and $$Pt,$$ which do not give a metal nitrate on treatment with concentrated $$HNO_3$$?

p - block Elements

When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide are liberated in equal volumes according to the equation: $$xCu+yHN{ O }_{ 3 }\longrightarrow Cu{ \left( N{ O }_{ 3 } \right) }_{ 2 }+NO+N{ O }_{ 2 }+{ H }_{ 2 }O$$ The coefficients of $$x$$ and $$y$$ are respectively:

p - block Elements

Illustrate how copper metal can give different products on reaction with HNO3.

p - block Elements

Concentrated nitric acid reacts with iodine to give:

p - block Elements

Write the formula of the compound of iodine which is obtained when $$conc. HNO_{3}$$ oxidises $$I_{2}$$.

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