You visited us 1 times! Enjoying our articles? Unlock Full Access!

Stoichiometric Calculations

You place 2...

Question

You place $2.56$ g $C a C O_{3}$ (calcium carbonate) in a beaker containing $250$ mL of $0.125$ M $H C l$ . What mass of $C a C l_{2}$ can be produced? (Answer in nearest integer)
$C a C O_{3} (s) + 2 H C l (a q) \to C a C l_{2} (a q) + C O_{2} (g) + H_{2} O (l)$

Open in App

Solution

The balanced chemical reaction is $C a C O_{3} (1 m o l) + 2 H C l (2 m o l) \to C a C l_{2} (1 m o l) + C O_{2} + H_{2} O$ .
$0.0256$ moles of calcium carbonate will react with $0.0512$ mol $H C l$ .
$250$ mL of $0.125$ M $H C l$ corresponds to $0.03125$ moles. They react with $0.015625$ mol calcium carbonate.
Hence, $H C l$ is the limiting reagent.
Unreacted calcium carbonate is $0.0256 - 0.0156 = 0.01$ mol $= 1.0$ g.
The number of moles of calcium carbonate reacted is equal to the mole of calcium chloride formed.
Mass of calcium carbonate formed is $1.7344$ g and nearest integer is $2$ .

Suggest Corrections

Similar questions

H C l

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}

25

0.75

H C l

10

H C l

0.3

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 l_{2}

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?

H C l

C a C l_{2} and 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}

25

mL

0.75 M

HCI

H C I

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

C a C O_{3}

25 m L

0.75 M H C l

Join BYJU'S Learning Program