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Limiting Reagent or Reactant

Which of the ...

Question

Which of the following statements is/are correct?
$C S_{2} + 3 C l_{2} Δ - \to C C l_{4} + S_{2} C l_{2}$
$1.0$ g of $C S_{2}$ and $2.0$ g of $C l_{2}$ reacts.

A

0.714

g

C S_{2}

is used in the reaction.

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B

0.286

g

C S_{2}

is in excess.

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C

1.45

g of

C C l_{4}

is formed.

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D

0.8

g

C l_{2}

is in excess.

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Solution

The correct options are
A $0.286$ g $C S_{2}$ is in excess.
B $0.714$ g $C S_{2}$ is used in the reaction.
C $1.45$ g of $C C l_{4}$ is formed.
$n_{P_{4}} = \frac{1.24}{31 \times 4} = 0.01 m o l, n_{O_{2}} = \frac{8}{32} = 0.25 m o l e s$
In reaction (i), moles of $O_{2}$ required
$= (0.01 m o l P_{4}) (\frac{5 m o l O_{2}}{m o l P_{4}})$
$= 0.01 \times 5 = 0.05 m o l$
Since there is more $O_{2}$ present than required
a. Therefore, $P_{4}$ is the limiting quantity.
b. Wrong.
c. $[M w P_{4} O_{10} = 284]$
0.01 mol of $P_{4}$ produces=0.01 mol of $P_{4} O_{10}$
$= (0.01 m o l P_{4}) (\frac{1 m o l P_{4} O_{10}}{m o l P_{4} O_{10}}) (\frac{284 g P_{4} O_{10}}{m o l P_{4} O_{10}})$
$= 0.01 \times 284 = 2.84 g$
Hence, (c) is wrong.
d. [Mw of $P_{4} O_{6} = 220$ ]
Weight of $P_{4} O_{6}$ produced
$= (0.01 m o l P_{4}) (\frac{1 m o l P_{4} O_{6}}{m o l P_{4} O_{6}}) (\frac{220 g P_{4} O_{6}}{m o l P_{4} O_{6}})$
$= 0.01 \times 220 = 2.2 g$
Hence, (d) is wrong.
Hence options A,B & C are correct.

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Similar questions

Q. Balance the following equation and choose the correct value of sum of coefficients of the products.

C S_{2} + C l_{2} \to C C l_{4} + S_{2} C l_{2}

.

Mention catalyst in the following reaction:

C S_{2} + 3 C l_{2} \to C C l_{4} + S_{2} C l_{2}

48^{\circ}

the vapour pressure of pure

C S_{2}

is 850 torr. A solution of 2.0 g of sulphur in 100 g of

C S_{2}

has a vapour pressure 844.9 trr. Determine the atomicity of sulphur molecule:

1400 K

, for the reaction

C H_{4} (g) + 2 H_{2} S (g) ⇌ C S_{2} (g) + 4 H_{2} (g)

K_{c} = 2.5 \times 10^{- 3}

10 L

reaction vessel contains 2 moles of

C H_{4},

C S_{2}

H_{2}

H_{2} S

. Choose the correct statement from the following:

Q. In the reaction

C S_{2} (l) + 3 O_{2} (g) \to C O_{2} (g) + 2 S O_{2} (g) △ H = - 265 k c a l

The enthalpies of formation of

C O_{2}

S O_{2}

are both negative and are in the ratio 4 : 3. The enthalpy of formation of

C S_{2}

is + 26 kcal/mol. Calculate the enthalpy of formation of

S O_{2} .

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