$2 V O + 3 F e_{2} O_{3} \to 6 F e O + V_{2} O_{5}$ 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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$F e_{2} O_{3}$

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$F e_{2} O$

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$V_{2} O_{5}$

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Solution

The correct option is B
$F e_{2} O_{3}$

If you check, the reaction is already balanced 2 moles of VO = 3 moles of $F e_{2} O_{3}$

$2 \times (47.87 + 16) = 3 \times (2 \times 55.85 + 3 \times 16)$

$2 (63.87) g = 3 \times (159.7) g$

$127.74 g$ of $V O$ = $478.1 g$ of $F e_{2} O_{3}$

$∴$ $2 g$ of $V O$ = $\frac{2 \times 478.1}{127.74} = 7.50 g$ of $F e_{2} O_{3}$

But only $5.75 g$ of $F e_{2} O_{3}$ is present.

This $F e_{2} O_{3}$ will get used up in the reaction and is the limiting reagent.

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Similar questions

$2 V O + 3 F e_{2} O_{3} \to 6 F e O + V_{2} O_{5}$ 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 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]

Q. 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

51

amu and of

F e

56

amu.)

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2VO + 3Fe2O3 → 6FeO + V2O5 If we start with 2g of VO and 5.75g of Fe2O3, 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 2g of VO and 5.75g of $F e_{2} O_{3}$ , which is the limiting reagent?

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