If $20.0 g$ of $C a C O_{3}$ is treated with $20.0 g$ of $H C l$ , how many grams of $C O_{2}$ can be obtained according to the following reaction:
$C a C O_{3} (s) + 2 H C l (a q) ⟶ C a C l_{2} (a q) + H_{2} O (l) + C O_{2} (g)$

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Solution

Moles of $C a C O_{3}$ = $\frac{20}{100} = 0.2$
$0.2$ moles of $C a C O_{3}$ would require $0.4$ moles $H C l$ .
Moles of $H C l$ = $\frac{20}{35.5} = 0.56$ (which is greater than $0.4$ )
$∴ C a C O_{3}$ is the limiting reagent.
$100 g$ of $C a C O_{3}$ produces $C O_{2} = 44 g$
$20 g$ of $C a C O_{3}$ will give $= \frac{44}{100} \times 20 = 8.8 g$ of $C O_{2}$

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Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, CaCO₃₍_s₎ + 2 HCl₍_aq₎ → CaCl₂₍_aq₎ + CO₂₍_g) + H₂O₍_l₎

What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?

C a C O_{3}

H C l

C O_{2}

$C a C O_{3} (s) + 2 H C l (a q)$

$↓$

$C a C l_{2} (a q) + H_{2} O (l) + C O_{2} (g)$

If 20.0g of CaCO₃ is treated with 20.0g of HCL how many grams of CO₂ will be produced ?

C a C O_{3}

C O_{2}

C a C O_{3} + 2 H C l \to C a C l_{2} + H_{2} O + C O_{2}

C a C O_{3} = 100 u, H C l = 36.5 u C O_{2} = 44 u)

C a C l_{2}

C O_{2}

C a C O_{3} (s) + 2 H C l (a q) \to C a C l_{2} (a q) + C O_{2} (g) + H_{2} O (l)

C a C O_{3}

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