Calculate the weight of FeO produced from 6-7 g of VO and 4-8 g of Fe2O3- 2VO-3Fe2O3- 6FeO-V2O5Given that the atomic weight of V is 51 amu and of Fe is 56 amu-

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Mole Fraction

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Question

Calculate the weight of $F e O$ produced from $6.7$ g of $V O$ and $4.8$ g of $F e_{2} O_{3}$ .
$2 V O + 3 F e_{2} O_{3} \to 6 F e O + V_{2} O_{5}$
(Given that the atomic weight of $V$ is $51$ amu and of $F e$ is $56$ amu.)

4.32

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7.755

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2.585

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0.0718

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Solution

The correct option is A $4.32$
Molar mass of $V O = 51 + 16 = 67 g / m o l$
Molar mass of $F e_{2} O_{3} = 2 (56) + 3 (16) = 160 g / m o l$
$6.7$ g $V O = \frac{6.7}{67} = 0.1 m o l$
$4.8$ g $F e_{2} O_{3} = \frac{4.8}{160} = 0.03$ mol
$0.1$ moles of VO will react with $0.1 \times \frac{3}{2} = 0.15$ moles of $F e_{2} O_{3}$ , however, only $0.03$ moles are present. Hence, $V O$ is excess reagent and $F e_{2} O_{3}$ is limiting reagent.
$0.03$ moles of $F e_{2} O_{3}$ will give $0.03 \times \frac{6}{3} = 0.06$ moles of $F e O$ .
Molar mass of $F e O = 56 + 16 = 72$ g/mol
Mass of $F e O = 72 \times 0.06 = 4.32$

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Similar questions

Q. Calculate the weight of

V_{2} O_{5}

produced from

5.75

g of

F e_{2} O_{3}

V O + F e_{2} O_{3} \to F e O + V_{2} O_{5}

(Fe

= 56

, V

= 51

, O

= 16

)

$2 V O + 3 F e_{2} O_{3} \to 6 F e O + V_{2} O_{5}$ .
If we start with 2 g of $V O$ and 6 g of $F e_{2} O_{3}$ , is the limiting reagent. [V = 48 g] [Fe = 56 g]

$2 V O + 3 F e_{2} O_{3} \to 6 F e O + V_{2} O_{5}$ If we start with 2g of VO and 5.75g of $F e_{2} O_{3}$ , which is the limiting reagent?

[V = 47.87g] [Fe = 55.85g]

$2 V O + 3 F e_{2} O_{3} \to 6 F e O + V_{2} O_{5}$ .
If we start with 2 g of $V O$ and 6 g of $F e_{2} O_{3}$ , is the limiting reagent. [V = 48 g] [Fe = 56 g]

$2 V O + 3 F e_{2} O_{3} \to 6 F e O + V_{2} O_{5}$ If we start with 2g of VO and 5.75g of $F e_{2} O_{3}$ , which is the limiting reagent?

[V = 47.87g] [Fe = 55.85g]

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Calculate the weight of FeO produced from 6.7 g of VO and 4.8 g of Fe2O3.2VO+3Fe2O3→6FeO+V2O5(Given that the atomic weight of V is 51 amu and of Fe is 56 amu.)

Calculate the weight of $F e O$ produced from $6.7$ g of $V O$ and $4.8$ g of $F e_{2} O_{3}$ .
$2 V O + 3 F e_{2} O_{3} \to 6 F e O + V_{2} O_{5}$
(Given that the atomic weight of $V$ is $51$ amu and of $F e$ is $56$ amu.)