
Which of the following statements are true?
(1) Cold and very dilute $HN{O_3}$ forms $N{H_4}N{O_3}$ with Zn or Sn
(2) Concentrated $HN{O_3}$ forms ${H_2}Sn{O_3}$ with Sn
(3) Cold and more concentrated $HN{O_3}$ forms $N{O_2}$ with Cu
(4) All the above are correct
Answer
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Hint: The given reactions of nitric acid depending on whether nitric acid is present in concentrated form or dilute form depending upon the concentration with different metals. Zinc reacts with dilute nitric acid and releases nitric oxide and cold and concentrated nitric acid releases nitrogen dioxide gas. Therefore, we will study the products in different reactions and get the right answer.
Complete step by step solution:
We will check for every option in the given question
The reaction of zinc with very dilute nitric acid: Zinc reacts with cold and very dilute nitric acid to form zinc nitrate nitrous oxide and water. The reaction is given down below as follows
$4Zn + 10HN{O_3} \to 4Zn{(N{O_3})_2} + 5{H_2}O + {N_2}O$
Here the products formed are zinc nitrate, nitric oxide, and water,
Therefore, option A is incorrect
The reaction of tin with concentrated nitric acid: Tin(Sn) reacts with concentrated nitric acid the products formed are meta stannic acid, nitrogen dioxide, and water. The reaction is given down as follows
$Sn + HN{O_3} \to {H_2}Sn{O_3} + N{O_2} + {H_2}O$
Here precipitate of meta stannic acid is removed. As on heating meta stannic acid, it converts to stannic oxide. The reaction is as follows
${H_2}Sn{O_3} \to Sn{O_2} + {H_2}O$${H_2}Sn{O_3} \to Sn{O_2} + {H_2}O$(On heating)
Hence, option B is incorrect
The reaction of copper and more concentrated nitric acid: Copper reacts with more concentrated nitric acid to form copper nitrate, nitrogen dioxide gas, and water.
The reaction is given down as follows
$4HN{O_3} + Cu \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O$
So, the correct answer is Option C.
Note: The metal reacts with an acid to form salt and hydrogen gas. The reaction is as follows
$Metal + Acid = Salt + Hydrogen(gas)$
Here nitric acid being a strong oxidizing agent oxidizes hydrogen into water.
It attacks most of the metals and oxidizes nonmetals and their compounds.
Complete step by step solution:
We will check for every option in the given question
The reaction of zinc with very dilute nitric acid: Zinc reacts with cold and very dilute nitric acid to form zinc nitrate nitrous oxide and water. The reaction is given down below as follows
$4Zn + 10HN{O_3} \to 4Zn{(N{O_3})_2} + 5{H_2}O + {N_2}O$
Here the products formed are zinc nitrate, nitric oxide, and water,
Therefore, option A is incorrect
The reaction of tin with concentrated nitric acid: Tin(Sn) reacts with concentrated nitric acid the products formed are meta stannic acid, nitrogen dioxide, and water. The reaction is given down as follows
$Sn + HN{O_3} \to {H_2}Sn{O_3} + N{O_2} + {H_2}O$
Here precipitate of meta stannic acid is removed. As on heating meta stannic acid, it converts to stannic oxide. The reaction is as follows
${H_2}Sn{O_3} \to Sn{O_2} + {H_2}O$${H_2}Sn{O_3} \to Sn{O_2} + {H_2}O$(On heating)
Hence, option B is incorrect
The reaction of copper and more concentrated nitric acid: Copper reacts with more concentrated nitric acid to form copper nitrate, nitrogen dioxide gas, and water.
The reaction is given down as follows
$4HN{O_3} + Cu \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O$
So, the correct answer is Option C.
Note: The metal reacts with an acid to form salt and hydrogen gas. The reaction is as follows
$Metal + Acid = Salt + Hydrogen(gas)$
Here nitric acid being a strong oxidizing agent oxidizes hydrogen into water.
It attacks most of the metals and oxidizes nonmetals and their compounds.
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