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Standard XII

Chemistry

Gay Lussac's Law, Avagadro's Law

Iron reacts w...

Question

Iron reacts with oxygen to form $F e_{2} O_{3}$ . If the atomic mass of iron is 56, the mass of oxygen that reacts with 2.5 moles of iron is

30.0g

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60.0g

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90.0g

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48.0g

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Solution

The correct option is B
60.0g

$4 F e + 3 O_{2} \to 2 F e_{2} O_{3}$

$4 m o l e o f F e (4 \times 56 = 224 g) r e a c t s w i t h \to 3 m o l e O_{2} (3 \times 32 = 96 g)$

$2.5 m o l e o f F e (2.5 \times 56 = 140 g) r e a c t s w i t h \to \frac{140}{224} \times 96 = 60 g o f O_{2}$

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Similar questions

Iron reacts with oxygen to form $F e_{2} O_{3}$ . If the atomic mass of iron is 56, the mass of oxygen that reacts with 2.5 moles of iron is

Q. When iron (Fe, gram atomic mass=56 g) is exposed to the atmospheric oxygen, it results in the formation of rust (

{Fe}_{2} O_{3}

, gram molecular mass=160 g). If 1120 g of

{Fe}_{2} O_{3}

is formed, how many moles of

{Fe}_{2} O_{3}

is formed?

I

. When iron reacts with oxygen an iron oxide compound with the formula of

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

can be formed.

I I

. Iron atoms form cations with a charge of

+ 3

by losing

3

electrons and oxygen atoms gain

2

electrons to form an anion with a

- 2

charge.

Q. During corrosion, iron reacts with oxygen and moisture present in the atmosphere to form rust.

2 H_{2} S (g) + 3 O_{2} (g) \to 2 S O_{2} (g) + 2 H_{2} O (g) +

heat

F e_{2} O_{3} (s) + 3 C O (g) \to 2 F e (s) + 3 C O_{2} (g)

According to the above reaction , 3 moles of $F e_{2} O_{3}$ are allowed react with 56 grams of CO so, find the approximately moles of iron produced ?

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Iron reacts with oxygen to form Fe2O3. If the atomic mass of iron is 56, the mass of oxygen that reacts with 2.5 moles of iron is

The correct option is B 60.0g 4Fe+3O2→2Fe2O3 4 mole of Fe(4×56=224g) reacts with→3 mole O2(3×32=96g) 2.5 mole of Fe(2.5×56=140g) reacts with→140224×96=60g of O2

Iron reacts with oxygen to form $F e_{2} O_{3}$ . If the atomic mass of iron is 56, the mass of oxygen that reacts with 2.5 moles of iron is

The correct option is B
60.0g

$4 F e + 3 O_{2} \to 2 F e_{2} O_{3}$

$4 m o l e o f F e (4 \times 56 = 224 g) r e a c t s w i t h \to 3 m o l e O_{2} (3 \times 32 = 96 g)$

$2.5 m o l e o f F e (2.5 \times 56 = 140 g) r e a c t s w i t h \to \frac{140}{224} \times 96 = 60 g o f O_{2}$