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Stoichiometric Calculations

Which of the ...

Question

Which of the following statements is/are correct?
$1.0$ g mixture of $C a C O_{3} (s)$ and glass beads liberate $0.22$ g of $C O_{2}$ upon treatment with excess of $H C l$ . Glass does not react with $H C l$ .
$C a C O_{3} + 2 H C l \to C O_{2} + H_{2} O + C a C l_{2}$
[Molecular weight of $C a C O_{3} = 100$ g/mol, $C O_{2} = 44$ g/mol and atomic weight of $C a = 40$ g]

The weight of

C a C O_{3}

in the original mixture is

0.5

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Tile weight of calcium in the original mixture is

0.2

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The weight percent of calcium in the original mixture is

40 % C a

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The weight percent of calcium in the original mixture is

20 % C a

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Solution

The correct options are
A The weight of $C a C O_{3}$ in the original mixture is $0.5$ g.
B Tile weight of calcium in the original mixture is $0.2$ g.
C The weight percent of calcium in the original mixture is $20 % C a$ .
Weight of $C a C O_{3} = \frac{0.22 \times 100}{44} = 0.5$ g $C a C O_{3}$
Moles of $C a C O_{3} =$ Moles of $C a$ $= (\frac{0.22}{44}) = 0.005$ mol
Weight of $C a = 0.005 \times 40 = 0.2$ g $C a$
Percentage of $C a$ $= \frac{0.2}{1.0} \times 100 = 20 %$ of $C a$

Suggest Corrections

Similar questions

10

g of a sample of a mixture of

C a C l_{2}

and

N a C l

is treated to precipitate all the calcium as

C a C O_{3}

. This

C a C O_{3}

is heated to convert all the

C a

C a O

and the final mass of

C a O

1.62

g. The percent by mass of

C a C l_{2}

in the original mixture is :

Q. A

10.0

g sample of a mixture of

C a C l_{2}

and

N a C l

is treated to precipitate all the calcium as calcium carbonate. This

C a C O_{3}

is heated to convert all the

C a

C a O

and the final mass of

C a O

1.62

g. If

X

percentage by mass of

C a C l_{2}

is in the original mixture, then value of

10 X

is :

Q. A Sample of mixture of

C a C l_{2}

and

N a C l

weighing

4.44 g

was treated to precipitate all the

C a

C a C O_{3}

, which was then heated and quantitatively converted to

1.12 g

C a O

. (Atomic weight: Ca= 40, Na= 23, Cl= 35.5)

Q. A

10.0 g

sample of a mixture of calcium chloride and sodium chloride is treated with

N a_{2} C O_{3}

to precipiate the calcium as calcium carbonate. This

C a C O_{3}

is heated to convert all the calcium to

C a O

and the final mass of

C a O

1.62 g

. The

%

by mass of

C a C l_{2}

in the original mixture is

Q. When the calcium carbonate is reacted with an excess amount of hydrochloric acid, how much

C a C O_{3}

is required to produce

11.2

liters of

C O_{2}

at STP ?

(Given molar masses

:

C a C O_{3} = 100

g/mol,

H C l = 36.5

g/mol,

C O_{2} = 44.0

g/mol)

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Which of the following statements is/are correct?1.0 g mixture of CaCO3(s) and glass beads liberate 0.22 g of CO2 upon treatment with excess of HCl. Glass does not react with HCl.CaCO3+2HCl→CO2+H2O+CaCl2[Molecular weight of CaCO3=100 g/mol, CO2=44 g/mol and atomic weight of Ca=40 g]