Calculate the weight of $F e O$ produced from $2 g V O$ and $5.75 g$ of $F e_{2} O_{3} .$ Also report the limiting reagent.

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

Open in App

Solution

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

Mole before reaction $2 / 67$ $5.75 / 160$

$= 0.0298$ $0.0359$

Mole after reaction $[0.0298 - (0.0359 \times 2) / 3] 0$ $[0.0359 \times 2] [0.0359 \times 1]$

$∵$ Mole ratio in reaction is $V O : F e_{2} O_{3} : F e O : V_{2} O_{5} :: 2 : 3 : 6 : 1$

$∴$ Mole of $F e O$ formed $= 0.0359 \times 2$

$∴$ Weight of $F e O$ formed $= 0.0359 \times 2 \times 72 = 5.17 g$

The limiting reagent is one which is totally used i.e, $F e_{2} O_{3}$ here.

Suggest Corrections

Similar questions

V_{2} O_{5}

5.75

F e_{2} O_{3}

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

= 56

= 51

= 16

F e O

6.7

V O

4.8

F e_{2} O_{3}

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

V

51

F e

56

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

V_{2} O_{5}

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

Join BYJU'S Learning Program