Question

Write the following reactions:

(i) Reaction between NaOH and Zn

(ii) Reaction between Potassium hydroxide and sulphuric acid​

Answer 1

Answer:

i) Zn + 2NaOH. Na2ZnO2 + H. 2

Zinc react with sodium hydroxide to produce zincate sodium and hydrogen. This reaction takes place at a temperature near 550°C.

ii) This means that: the reactants are potassium hydroxide and sulfuric acid. the products are potassium sulfate and water. the word equation is: potassium hydroxide + sulfuric acid → potassium sulfate + water.

Explanation:

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Answer 2

Answer:

$\rule{300}{1.5}$

i) Reaction b/w NaOH and Zn:

• Sodium hydroxide (NaOH) reacts with Zinc (Zn) to give a complex salt called Sodium zincate.
• The process is very slow.
• The process evolves Hydrogen (H₂) gas as a by product.
• The presence of hydrogen gas is determined by bringing a lighted matchstick which puts off due to Hydrogen gas.

Chemical equation:

• $\displaystyle\sf 2\;NaOH_{\;(aq)}+Zn_{\;(s)}\displaystyle\overset{\Delta}{\longrightarrow} Na_{2}ZnO_{2}_{\;(aq)}+H_{2}_{\;(g)}\uparrow$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

ii) Reaction b/w KOH and sulphuric acid:​

• Potassium hydroxide (KOH) reacts with sulphuric acid (H₂SO₄) to give Potassium sulphate.
• Along with the salt Water molecule is liberated.
• This is a good example of Neutralization reaction.
• And, it is also an example of Double - displacement reaction.

Chemical equation:

• $\sf 2\;KOH_{\;(aq)}+H_{2}SO_{4}_{\;(aq)}\longrightarrow K_{2}SO_{4}_{\;(aq)}+2\;H_{2}O$

$\rule{300}{1.5}$