When FeS2 is burnt in air- it converts to Fe2O3 as- FeS2s - O2g - Fe2O3s - SO2g- the change in percentage by weight of iron in the process is- Atomic mass of Fe - 56 u

FeS_2 (s)+ O_2 (g)→ Fe_2O_3(s) + SO_2(g) For FeS_2, Molecular weight = 56 + 32 × 2= 120 u ∴ % of Fe = 56120× 100 = 46.67% For Fe_2O_3, Molecular weight= 56 × 2 ...

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Byju's Answer

Standard XII

Chemistry

Laws of Mass Conservation

When FeS2 is ...

Question

When $F e S_{2}$ is burnt in air, it converts to $F e_{2} O_{3}$ as, $F e S_{2} (s) + O_{2} (g) \to F e_{2} O_{3} (s) + S O_{2} (g)$ , the change in percentage by weight of iron in the process is: $(Atomic mass of Fe = 56 u)$

23 %

increase

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12 %

decrease

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12 %

increase

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No change

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Solution

The correct option is A $23 %$ increase
$F e S_{2} (s) + O_{2} (g) \to F e_{2} O_{3} (s) + S O_{2} (g)$
For $F e S_{2},$
$Molecular weight = 56 + 32 \times 2 = 120 u$
$∴ % of Fe = \frac{56}{120} \times 100 = 46.67 %$
For $F e_{2} O_{3},$
$Molecular weight = 56 \times 2 + 3 \times 16 = 160 u$
$∴ % of Fe = \frac{112}{160} \times 100 = 70 %$
$Percentage increase = 70 - 46.67 = 23.33 %$

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Similar questions

Q. Equivalent weight of

F e S_{2}

in the half reaction,

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

is:

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. Equivalent weight of a aFeS2 is in the half reaction
FeS2--->Fe2O3+SO2

Q. 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

F e S_{2}

is used.

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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When FeS2 is burnt in air, it converts to Fe2O3 as, FeS2(s)+O2(g)→Fe2O3(s)+SO2(g), the change in percentage by weight of iron in the process is: (Atomic mass of Fe = 56 u)

The correct option is A 23% increaseFeS2(s)+O2(g)→Fe2O3(s)+SO2(g) For FeS2, Molecular weight=56+32×2=120 u ∴ % of Fe=56120×100=46.67% For Fe2O3, Molecular weight=56×2+3×16=160 u ∴ % of Fe=112160×100=70% Percentage increase=70−46.67=23.33%

When $F e S_{2}$ is burnt in air, it converts to $F e_{2} O_{3}$ as, $F e S_{2} (s) + O_{2} (g) \to F e_{2} O_{3} (s) + S O_{2} (g)$ , the change in percentage by weight of iron in the process is: $(Atomic mass of Fe = 56 u)$