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Stoichiometric Calculations

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Question

Find out the weight of calcium carbonate that must be decomposed to produce sufficient quantity of carbon dioxide to convert 10.6gm of sodium carbonate completely into sodium bocarbonate.

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Solution

$C a C O_{3} + H_{2} O + {N a}_{2} {C O}_{3} ⟶ C a O + N a H {C O}_{3}$
Mw of $C a C O_{3} = 100 {N a}_{2} {C O}_{3} = 10.6.$

$10.6 g m = 100 \times \frac{10.6}{10.6} = 10 g m o f C a {C O}_{3}$
Therefore $10.6 g m = 100 \times \frac{10.6}{10.6} = 10 g m o f C a {C O}_{3} .$

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0.5

C a C O_{3} ⟶ C a O + C O_{2}

N a_{2} C O_{3} + H_{2} O + C O_{2} ⟶ 2 N a H C O_{3}

0.5

C a C O_{3} ⟶ C a O + C O_{2}

N a_{2} C O_{3} + H_{2} O + C O_{2} ⟶ 2 N a H C O_{3}

C a C O_{3}

N a_{2} C O_{3}

N a H C O_{3}

N a = 23

C a = 40

C a C O_{3} \to C a O + C O_{2}

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

C a C O_{3}

21.2 g

N a_{2} C O_{3}

N a H C O_{3}

N a = 23 g/mol, C a = 40 g/mol

C a C O_{3} \to C a O + C O_{2}

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

C a C O_{3}

21.2 g

N a_{2} C O_{3}

N a H C O_{3}

N a = 23 g/mol, C a = 40 g/mol

C a C O_{3} \to C a O + C O_{2}

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

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