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Standard XII

Chemistry

Preparation of Benzene and Homologues

4 moles of ca...

Question

4 moles of calcium carbonate react with 7 moles of aqueous HCl to give $C a C l_{2}$ and $C O_{2}$ according to the given reaction:
$C a C O_{3} (s) + 2 H C l (a q) ⟶ C a C l_{2} (a q) + C O_{2} (g) + H_{2} O (l)$

Calculate the mass of $C a C l_{2}$ produced in the reaction.
(atomic weight: Ca = 40 u, C = 12 u, 0 = 16 u, Cl = 35.5 u)

388.50 g

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522.20 g

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222.50 g

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443.25 g

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Solution

The correct option is A 388.50 g
According to the reaction,
1 mole of $C a C O_{3}$ reacts with 2 moles of HCl.
So, 4 moles of $C a C O_{3}$ will react with 8 moles of HCl. Since we have 7 moles of HCl, HCl will be the limiting reagent.

2 moles of HCl produce 1 mole of $C a C l_{2}$ .
Thus, 7 moles of HCl will produce 3.5 moles of $C a C l_{2}$ .
Molar mass of $C a C l_{2} = 40 + 35.5 \times 2 = 111 g / m o l$
$∴$ mass of $C a C l_{2}$ produced,
= $111 g / m o l \times 3.5 m o l$
$= 388.5 g$

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Similar questions

Q. Calcium carbonate reacts with aqueous HCl to give

C a C l_{2}

and

C O_{2}

according to the reaction:

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

When 250 mL of 0.76 M HCl reacts with 1000 g of

C a C O_{3}

, calculate the number of moles of

C a C l_{2}

formed:

Calcium carbonate reacts with aqueous HCl to give $C a C l_{2}$ and $C O_{2}$ according to the reaction given below

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

What mass of $C a C l_{2}$ will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of $C a C O_{3}$ ? Name the limiting reagent. Calculate the number of moles of $C a C l_{2}$ formed in the reaction.

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?

Q. Calcium carbonate (

10

g) reacts with aqueous

H C l

(

0.3

mol) according to the reaction,

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)

The number of moles of

C a C l_{2}

formed in the reaction is:

Q. Calcium carbonate reacts with aqueous HCl to give

C a C l_{2}

and

C O_{2}

according to the reaction -

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)

The mass of

C a C O_{3}

which is required to react completely with 25 ml of 0.75 M HCl is:

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4 moles of calcium carbonate react with 7 moles of aqueous HCl to give CaCl2 and CO2 according to the given reaction: CaCO3(s)+2HCl(aq)⟶ CaCl2(aq)+CO2(g)+H2O(l) Calculate the mass of CaCl2 produced in the reaction. (atomic weight: Ca = 40 u, C = 12 u, 0 = 16 u, Cl = 35.5 u)