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Alkali Metals

The product/s...

Question

The product/s of the reaction, is/are:
${N a}_{2} {C O}_{3} + {C O}_{2} + H_{2} O ⟶$

2 N a O H + {C O}_{2}

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{N a}_{2} {C O}_{3} + H_{2} {C O}_{3}

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2 N a H {C O}_{3}

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None of these

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Solution

The correct option is D $2 N a H {C O}_{3}$
When sodium carbonate is treated with ${C O}_{2}$ and $H_{2} O$ , it gets converted into sodium bicarbonate.

${N a}_{2} {C O}_{3} + {C O}_{2} + H_{2} O ⟶ 2 N a H C O_{3} s o d i u m b i c a r b o n a t e$

Suggest Corrections

Similar questions

{A l}_{2} O_{3} . {2 H}_{2} O + {N a}_{2} {C O}_{3} ⟶ {2 N a A l O}_{2} + {2 H}_{2} O + {C O}_{2}

{2 N a A l O}_{2} + {2 H}_{2} O + {C O}_{2} 50 - 60^{o} C - ------ \to {A l}_{2} O_{3} . {2 H}_{2} O + {N a}_{2} {C O}_{3}

N a H {C O}_{3}

N a H {C O}_{3} \to {N a}_{2} {C O}_{3} + H_{2} O + {C O}_{2}

0.50 m

{N a}_{2} {C O}_{3}

0.50 M

{N a}_{2} {C O}_{3}

250 m L

{N a}_{2} {C O}_{3}

2.65 g

{N a}_{2} {C O}_{3} \cdot 10 m L

x

m L

0.001 M

{N a}_{2} {C O}_{3}

x

{N a}_{2} {C O}_{3} = 106

(50 + x)

x

{N a}_{2} {C O}_{3} + H_{2} S O_{4} ⟶ {N a}_{2} S O_{4} + H_{2} O + {C O}_{2}

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