For the preparation of H2SO4 from iron pyrite -FeS2- involves following set of reaction-FeS2-O2 - Fe2O3-SO2SO2 -O2 - SO3SO3-H2O - H2SO4Calculate the weight of H2SO4 which can be obtained when 240 kg of FeS2 is used-

Applying POAC of S, 2 moles of FeS_2=1 × moles of H_2SO_4 2 ×[ weight of FeS_2MM of FeS_2 nbsp; ]=1 ×[ weight of H_2SO_4MM of H_2SO_4 nbsp; ] 2 ×[ 240 × ...

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Byju's Answer

Standard XII

Chemistry

Balancing Redox Equations

For the prepa...

Question

For the preparation of $H_{2} S O_{4}$ from iron pyrite $[F e S_{2}]$ involves following set of reaction.
$F e S_{2} + O_{2} \to F e_{2} O_{3} + S O_{2}$
$S O_{2} + O_{2} \to S O_{3}$
$S O_{3} + H_{2} O \to H_{2} S O_{4}$
Calculate the weight of $H_{2} S O_{4}$ which can be obtained when $240 kg$ of $F e S_{2}$ is used.

392 kg

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412 kg

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329 kg

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292 kg

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Solution

The correct option is A $392 kg$
Applying POAC of $S$ ,
$2 moles of F e S_{2} = 1 \times moles of H_{2} S O_{4}$
$2 \times [\frac{weight of F e S_{2}}{MM of F e S_{2}}] = 1 \times [\frac{weight of H_{2} S O_{4}}{MM of H_{2} S O_{4}}]$

$2 \times [\frac{240 \times 10^{3}}{MM of F e S_{2}}] = 1 \times [\frac{weight}{98}]$

$weight of H_{2} S O_{4} = 392 kg$

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Similar questions

Q. How many kilograms of pure

H_{2} S O_{4}

could be obtained from 1 kg of iron pyrites

(F e S_{2})

according to the following reactions?

4 F e S_{2} + 11 O_{2} \to 2 F e_{2} O_{3} + 8 S O_{2}

2 S O_{2} + O_{2} \to 2 S O_{3}

S O_{3} + H_{2} O \to H_{2} S O_{4}

Q. Equivalent weight of

F e S_{2}

[mol.wt = M] in the following reaction is?

F e S_{2} + O_{2} \to F e_{2} O_{3} + S O_{2}

Q. Calculate the mass of a coal sample in kg containing 0.05 % mass of iron pyrite

(F e S_{2})

that can produce 44.8 litres of

S O_{2}

at 1 atm and

273 K

with 40 % reaction yield.

F e S_{2} + O_{2} \to F e_{2} O_{3} + S O_{2}

(Molar mass of Iron =

56 g/mol

; molar mass of Sulphur =

32 g/mol

)

Q. Balance the following equation.

F e S_{2} + O_{2} \to F e_{2} O_{3} + S O_{2}

Q. '

n

' factor for

S O_{2}

F e S_{2} + O_{2} ⟶ F e_{2} O_{3} + S O_{2}

, is :

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For the preparation of H2SO4 from iron pyrite [FeS2] involves following set of reaction. FeS2+O2→Fe2O3+SO2 SO2+O2→SO3 SO3+H2O→H2SO4 Calculate the weight of H2SO4 which can be obtained when 240 kg of FeS2 is used.

The correct option is A 392 kgApplying POAC of S, 2 moles of FeS2=1× moles of H2SO4 2×[weight of FeS2MM of FeS2]=1×[weight of H2SO4MM of H2SO4] 2×[240×103MM of FeS2]=1×[weight 98] weight of H2SO4=392 kg