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Byju's Answer

Standard XII

Chemistry

Percentage Composition

4 g of hydrog...

Question

$4$ g of hydrogen $(H_{2})$ , $64$ g of sulphur (S) and $44.8$ L of $O_{2}$ at STP react and form $H_{2} S O_{4}$ . If $49$ g of $H_{2} S O_{4}$ is formed, then $%$ yield is ?

25 %

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50 %

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75 %

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100 %

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Solution

The correct option is B $50 %$
$H_{2} + S + 2 O_{2} ⟶ H_{2} S O_{4}$
$2 g$ $32 g$ $64 g | | |$ $98 g$
$44.8 L$
The mole ratio is $H : S : O = 1 : 1 : 2$
Since $O_{2}$ is limiting only $98 g$ of $H_{2} S O_{4}$ is formed.
Given $49 g$ of $H_{2} S O_{4}$ is formed.
So % yield= $\frac{49}{98} \times 100 = 50$ % .

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Similar questions

Q. Peroxydisulfuric acid

(H_{2} S_{2} O_{8})

can be prepared by electrolytic oxidation of

H_{2} S O_{4}

as:

2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} + 2 H^{+} + 2 e

O_{2}

and

H_{2}

are by products. In such on electrolysis

0.87

g of

H_{2}

and

3.36

O_{2}

were generated at STP. Report the mass of

H_{2} S_{2} O_{8}

formed in decagram (write nearest integer).

Q. Assertion (A):

109

H_{2} S O_{4}

represent a way to express concentration of industrial

H_{2} S O_{4}

.
Reason (R): It represents that

9

H_{2} O

reacts with

40

S O_{3}

to produce

49

H_{2} S O_{4}

in addition to

100

H_{2} S O_{4}

S (s) + O_{2} (g) \to S O_{2} (g) (Y i e l d = 80 %)

S O_{2} (g) + C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l (Y i e l d = 30 %)

80 g

of Sulphur and

112 L

of oxygen at STP is burnt to form

S O_{2},

which is oxidized by

56 L

C l_{2}

gas at STP dissolved in water. This solution is treated with

B a C l_{2}

solution. Calculate the moles of precipitate formed?
(Molar mass of barium =

137 g / m o l

)

S + O_{2} \to S O_{2} (Yield = 80%)

S O_{2} + C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l (Yield = 60%)

H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l (Yield = 30%)

8 g of Sulphur is burnt to form

S O_{2}

, which is oxidized by

C l_{2}

water. The solution is then treated with

B a C l_{2}

solution. The amount of

B a S O_{4}

precipitated is:
(Molar mass of Barium =

137 g/mol

)

Q. Number of g moles of sulphur present in 49 g of

H_{2} S O_{4}

are:

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4 g of hydrogen (H2), 64g of sulphur (S) and 44.8 L of O2 at STP react and form H2SO4. If 49g of H2SO4 is formed, then % yield is ?

The correct option is B 50%H2+S+2O2⟶H2SO4 2g 32g 64g||| 98g 44.8LThe mole ratio is H:S:O=1:1:2Since O2 is limiting only 98g of H2SO4 is formed.Given 49g of H2SO4 is formed.So % yield= 4998×100=50% .

$4$ g of hydrogen $(H_{2})$ , $64$ g of sulphur (S) and $44.8$ L of $O_{2}$ at STP react and form $H_{2} S O_{4}$ . If $49$ g of $H_{2} S O_{4}$ is formed, then $%$ yield is ?