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Mole Concept

Mole ratio of...

Question

Mole ratio of $F e$ in $F e O$ , ${F e}_{2} O_{3}$ and ${F e}_{3} O_{4}$ samples of equal mass is:

A

1 : 2 : 3

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B

0.9 : 1 : 0.93

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C

1 : 0.9 : 0.93

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D

3 : 2 : 1

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Solution

The correct option is B $1 : 0.9 : 0.93$
Let mass of all compounds is w gm, then mole ratio of Fe in $F e O$ , ${F e}_{2} O_{3}$ and ${F e}_{3} O_{4}$ is, $\frac{w}{72} : \frac{2 w}{160} : \frac{3 w}{232} = 1 : 0.9 : 0.93$ .

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Similar questions

Q. Three samples of iron equal weights are used to from FeO,

{F e}_{2} O_{3}

{F e}_{3} O_{4}

independently. Then, the amount (in grams)
of

O_{2}

Q. From the following data at 25C, Calculate the standard enthalpy of formation of FeO(s) and of

{F e}_{2} O_{3} (s)

.

	Reaction	$Δ_{r} H (K J / m o l e)$
(1)	${F e}_{2} O_{3} (s) + 3 C (g r a p h i t e) \to 2 F e (s) + C O (g)$	492
(2)	$F e O (s) + C (g r a p h i t e) \to F e (s) + C O (g)$	155
(3)	$C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g)$	-393
(4)	$C O (g) + 1 / 2 O_{2} (g) \to C O_{2} (g)$	-282

Δ G = - 177 K c a l

(1) 2 F e (s) + \frac{3}{2} O_{2} (g) ⟶ {F e}_{2} O_{3} (s)

Δ G = - 19 K c a l f o r (2) {4 F e}_{2} O_{3} (s) ⟶ {3 F e}_{3} O_{4} (s)

What is the Gibbs free energy of formation of

{F e}_{3} O_{4}

?

Q. Determine the heat of the following reaction:

F e O (s) + {F e}_{2} O_{3} (s) ⟶ {F e}_{3} O_{4} (s)

Given informations:

2 F e (s) + O_{2} (g) ⟶ 2 F e O (s); Δ H^{o} = - 544 k J

4 F e (s) + 3 O_{2} (g) ⟶ 2 {F e}_{2} O_{; 3} (s); Δ H^{o} = - 1648.4 k J

{F e}_{3} O_{4} (s) ⟶ 3 F e (s) + 2 O_{2} (g); Δ H^{o} = + 1118.4 k J

The mass percentage of ${F e}^{+ 3}$ ion present in $F e_{0.93} O_{1.0}$ is?

1)15%.

2) 5.5%.

3) 10%.

4) 11.5%

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