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Coordination Complexes

Ferric alum h...

Question

Ferric alum has the composition $(N H_{4})_{2} S O_{4} . F e_{2} (S O_{4})_{3} . x H_{2} O .$ Find the value of $x .$

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Solution

The correct option is C 6
Ferric
$(N H_{4})_{2} S O_{4} . F e_{2} (S O_{4})_{3} .24 H_{2} O$

Suggest Corrections

Similar questions

(N H_{4})_{2} S O_{4} . F e_{2} (S O_{4})_{3} .24 H_{2} O

F e

0.0056

25

F e_{2} (S O_{4})_{3} . (N H_{4})_{2} . S O_{4} .24 H_{2} O

1.25

F e + F e_{2} (S O_{4})_{3} ⟶ 3 F e S O_{4}

0.107 N

K M n O_{4}

F e

100

F e_{2} (S O_{4})_{3} . (N H_{4})_{2} S O_{4} .24 H_{2} O

= 964

2.41

F e

F e + F e_{2} (S O_{4})_{3} \to 3 F e S O_{4}

\frac{M}{60} K_{2} C r_{2} O_{7}

F e S O_{4}

C u

F e_{2} (S O_{4})_{3}

A l_{2} (S O_{4})_{3} . x H_{2} O

8.20

A l

x / 3

N H_{4} S C N

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