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Laws of Definite Proportions, Multiple Proportions

How many gram...

Question

How many grams of $C O_{2}$ can be prepared from $150$ grams of calcium carbonate ( $C a C O_{3}$ ) reacting with an excess of hydrochloric ( $H C l$ ) acid solution?

11

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22

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33

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44

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66

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Solution

The correct option is E $66$
$C a C O_{3} + 2 H C l \to C a C l_{2} + H_{2} O + C O_{2} ↑$
1 mole of calcium carbonate gives 1 mole of $C O_{2}$
Molar mass of calcium carbonate $= 40 + 12 + 3 (16) = 100$ g/mol.
150 grams of calcium carbonate corresponds to $\frac{150 g}{100 g/mol} = 1.50$ moles of calcium carbonate.
They will give 1.50 moles of $C O_{2}$ .
The molar mass of $C O_{2}$ is 44 g/mol.
Mass of $C O_{2}$ $= 1.5 mol \times 44 g/mol = 66$ mol

Suggest Corrections

Similar questions

C a C O_{3}

11.2

C O_{2}

:

C a C O_{3} = 100

H C l = 36.5

C O_{2} = 44.0

{C O}_{2}

100

{C a C O}_{3}

H C l

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}

M n O_{2}

M n O_{2} (s) \to 2 H_{2} O (l) + M n C l_{2} (a q) + C l_{2} (g)

H C l

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