Calculate the weight of V2O5 produced from 5-75 g of Fe2O3- VO-Fe2O3- FeO-V2O5Fe -56- V -51- O -16

Reaction Involved 2VO + 3Fe_2O_3→ #160; 6FeO + V_2O_5 Molecular mass of VO = 66.94 g/mol #160;Mass of 2 moles of VO #160; = 133.88 g #160;Molecular ...

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Standard XII

Chemistry

Percentage Composition

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Question

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 $= 16$ )

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Solution

Reaction Involved-
$2 V O + 3 F e_{2} O_{3} \to 6 F e O + V_{2} O_{5}$
Molecular mass of VO = 66.94 g/mol
Mass of 2 moles of VO = 133.88 g
Molecular mass of $F e_{2} O_{3}$ = 159.69 g/mol
Mass of 3 moles = 479 g
Thus,
133.88 gof VOrequires479 g $F e_{2} O_{3}$
2 g of VO requires $F e_{2} O_{3}$ = $\frac{479}{133.88}$ x 2 = 7.155 g
But we have only 5.75 g $F e_{2} O_{3}$ _.
Hence, $F e_{2} O_{3}$ is the limiting reagent
Molecular mass of FeO = 71.84 g/mol
Molecular mass of 6 mol FeO = 431.04 g
From 479 g $F e_{2} O_{3}$ we are getting 431.04 g FeO
Therefore, from 5.75 g, FeO formed = $\frac{431.04}{479}$ x 5.75
= 5.17 g

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Similar questions

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.)

Q. In the reaction

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

, what is the n-factor for

V_{2} O_{5}

Q. In the reaction,

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

, the equivalent mass of

V_{2} O_{5}

is equal to its :

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

F e O

produced from

2

V O

and

5.75

g of

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

is (as nearest integer) :

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

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Calculate the weight of V2O5 produced from 5.75 g of Fe2O3.VO+Fe2O3→FeO+V2O5(Fe=56, V=51, O=16)

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 $= 16$ )