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Standard XII

Chemistry

Stoichiometric Calculations

Al and ...

Question

$A l$ and $K C l O_{3}$ react together to form $A I_{2} O_{3}$ according to the following reaction:
$2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}$
$4 A l + 3 O_{2} ⟶ 2 A l_{2} O_{3}$
$4$ moles of $K C l O_{3}$ ( $50 %$ pure) on reaction with excess of $A l$ forms $A l_{2} O_{3}$ equal to :

2

mol

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4

mol

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6

mol

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8

mol

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Solution

The correct option is A $2$ mol
The reactions are as follows:
$2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}$
$4 A l + 3 O_{2} ⟶ 2 A l_{2} O_{3}$
$50 %$ of $4$ moles of $2 K C l O_{3}$ means $2$ moles.
According to the balanced reaction, $2$ moles of $K C l O_{3}$ forms $2$ moles of $A l_{2} O_{3}$ .

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Similar questions

A l

and

K C l O_{3}

react together to form

A l_{2} O_{3}

according to the following reactions :

2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}

4 A l + 3 O_{2} ⟶ 2 A I_{2} O_{3}

4

moles of

K C l O_{3}

(

50 %

pure) on reaction with excess of Al forms _______ moles of

A l_{2} O_{3}

Q. Al and

K C l O_{3}

react together to form

A I_{2} O_{3}

according to

2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}

4 A l + 3 O_{2} ⟶ 2 A I_{2} O_{3}

4 m o l e s o f K C l O_{3}

(50% pure) on reaction with excess of

A l

forms

x

moles of

A I_{2} O_{3}

A l

and

K C l O_{3}

react together to form

A l_{2} O_{3}

as follows:

2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}

4 A l + 3 O_{2} ⟶ 2 A l_{2} O_{3}

4

moles of

K C l O_{3}

(

50 %

pure) on reaction with excess of

A l

forms x moles of

A l_{2} O_{3}

. The value of x (as nearest integer) is:

5.0

g of

K C l O_{3}

gave

0.03

mol of

O_{2}

. Hence, percentage purity of

K C l O_{3}

is :

2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}

Q. Consider the following set of reactions:

A l + 6 H B r \to A l B r_{3} + H_{2} C_{2} H_{2} + H_{2} \to C_{2} H_{4}

Find the number of moles of ethylene produced when 81 g of aluminium and 40.5 g of HBr reacts to form hydrogen which in turn reacts with 1 g of

C_{2} H_{2}

to form

C_{2} H_{4}

.
(Molar mass of Al = 27 g/mol)

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Al and KClO3 react together to form AI2O3 according to the following reaction:2KClO3⟶2KCl+3O24Al+3O2⟶2Al2O34 moles of KClO3 (50% pure) on reaction with excess of Al forms Al2O3 equal to :

The correct option is A 2 molThe reactions are as follows:2KClO3⟶2KCl+3O24Al+3O2⟶2Al2O350% of 4 moles of 2KClO3 means 2 moles. According to the balanced reaction, 2 moles of KClO3 forms 2 moles of Al2O3.

$A l$ and $K C l O_{3}$ react together to form $A I_{2} O_{3}$ according to the following reaction:
$2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}$
$4 A l + 3 O_{2} ⟶ 2 A l_{2} O_{3}$
$4$ moles of $K C l O_{3}$ ( $50 %$ pure) on reaction with excess of $A l$ forms $A l_{2} O_{3}$ equal to :

The correct option is A $2$ mol
The reactions are as follows:
$2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}$
$4 A l + 3 O_{2} ⟶ 2 A l_{2} O_{3}$
$50 %$ of $4$ moles of $2 K C l O_{3}$ means $2$ moles.
According to the balanced reaction, $2$ moles of $K C l O_{3}$ forms $2$ moles of $A l_{2} O_{3}$ .