A certain mixed oxide of iron contains $2.5 g$ of oxygen for every $7 g$ of iron. If it is regarded as a mixture of $F e O$ and $F e_{2} O_{3}$ in the weight ratio $a : b$ , what is $a : b$ ?
(atomic weight of iron = 56)

9 : 10

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9 : 20

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14 : 5

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1 : 1

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Solution

The correct option is A $9 : 10$
Consider $x$ mol of $F e O$ and $y$ mol of $F e_{2} O_{3}$ to be present.

$\begin{matrix} F e O \to & F e + & O x & x & x F e_{2} O_{3} \to & 2 F e + & 3 O y & 2 y & 3 y \end{matrix}$

Given: weight ratio, $O : F e = 2.5 : 7$
Therefore,
$\frac{(x + 3 y) 16}{(x + 2 y) 56} = \frac{2.5}{7}$
$\Rightarrow \frac{x + 3 y}{x + 2 y} = \frac{2.5}{7} \times \frac{56}{16} = \frac{2.5}{2}$
Solving this, we get $\frac{x}{y} = \frac{2}{1}$
$\Rightarrow wt. ratio, \frac{F e O}{F e_{2} O_{3}} = \frac{2}{1} \times \frac{72}{160} = \frac{9}{10}$

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A certain oxide of iron contains 2.5 g of oxygen for every 7.0 g of iron. If it is regarded as mixture of FeO and $F e_{2} O_{3}$ in the weight ratio x : y, what is x : y, (atomic weight of iron = 56)

Q. The haemoglobin contains

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67200

. What is the number of iron atoms in each molecule (atomic weight of iron

= 56

1 m o l

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(F e)

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0.65 m o l O_{2}

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F e O

and

F e_{2} O_{3}

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F e_{2} O_{3}

would be:

(F e = 56)

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0.1210

g. Hence, formula of the iron oxide is

(F e = 56)

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A certain mixed oxide of iron contains 2.5 g of oxygen for every 7 g of iron. If it is regarded as a mixture of FeO and Fe2O3 in the weight ratio a:b, what is a:b? (atomic weight of iron = 56)

The correct option is A 9:10Consider x mol of FeO and y mol of Fe2O3 to be present. FeO→Fe+OxxxFe2O3→2Fe+3Oy2y3y Given: weight ratio, O:Fe=2.5:7 Therefore, (x+3y)16(x+2y)56=2.57 ⇒x+3yx+2y=2.57×5616=2.52 Solving this, we get xy=21 ⇒wt. ratio,FeOFe2O3=21×72160=910

A certain mixed oxide of iron contains $2.5 g$ of oxygen for every $7 g$ of iron. If it is regarded as a mixture of $F e O$ and $F e_{2} O_{3}$ in the weight ratio $a : b$ , what is $a : b$ ?
(atomic weight of iron = 56)