CameraIcon

SearchIcon

MyQuestionIcon

1

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

Stoichiometric Calculations

S + O2 → SO2 ...

Question

$S + O_{2} \to S O_{2} (Yield = 80%)$
$S O_{2} + C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l (Yield = 60%)$
$H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l (Yield = 30%)$
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 = $137 g/mol$ )

A

8.39 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

B

18.39 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

24.26 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

13.67 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $8.39 g$
$S + O_{2} \to S O_{2}$ $(I) (Yield = 80%)$
$S O_{2} + C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l$ $(I I) (Yield = 60%)$
$H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l$ $(I I I) (Yield = 30%)$
In reaction (i),
1 mol of $S$ forms 1 mol of $S O_{2}$
Again, $8 g$ sulphur = $\frac{8}{32}$ mol of $S$ = 0.25 mol $S$
Yield of the reaction is given as 80 %
Actual moles of $S O_{2}$ formed $= 0.25 \times 0.80 = 0.2 m o l$
In reaction (ii),
1 mol of $S O_{2}$ forms 1 mol of $H_{2} S O_{4}$
0.2 mol of $S O_{2}$ will form 0.2 mol of $H_{2} S O_{4}$
Yield of the reaction is given as 60 %
Actual amount of $H_{2} S O_{4}$ formed $= 0.2 \times 0.60 = 0.12 m o l$
In reaction (iii),
1 mol of $H_{2} S O_{4}$ reacts to form 1 mol of $B a S O_{4}$
0.12 mol of $H_{2} S O_{4}$ reacts to form 0.12 mol of $B a S O_{4}$
Yield of the reaction is given as 30 %
So moles of $B a S O_{4}$ produced = $0.12 \times 0.3 = 0.036 m o l$
Molar mass of $B a S O_{4} = 137 + 32 + 64 = 233 g$
Actual amount of precipitate obtained $0.036 \times 233 g = 8.39 g$

Suggest Corrections

thumbs-up

0

Similar questions

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

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

)

8 g

of sulphur are burnt to form

S O_{2}

which is oxidized by

C l_{2}

water. The solution is treated with

B a C l_{2}

solution. The amount of

B a S O_{4}

precipitated is :

8

g of sulfur is burnt to form

S O_{2}

which is oxidized by

C l_{2}

water. The solution is treated with

B a C l_{2}

solution. Moles of

B a S O_{4}

precipitated are:

8 g of sulphur is burnt to form $C l_{2}$ which is oxidized by $C l_{2}$ water. The solution is treated with $B a C l_{2}$ solution. Number of moles of $B a S O_{4}$ precipitated is

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Stoichiometric Calculations

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Stoichiometric Calculations

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon