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Byju's Answer

Standard XII

Chemistry

Concentration Terms - An Introduction (w/w etc)

Al and KClO...

Question

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 $A I_{2} O_{3}$ .

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
$2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}$
$4 A l + 3 O_{2} ⟶ 2 A l_{2} O_{3}$
50% 4 moles means 2 mole, so, according to balance reaction it can be seen that 2 mole 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 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 :

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}

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. What volume of oxygen at STP can be produced by 6.125 g of potassium chlorate according to the reaction?

2 K C l O_{3} \to 2 K C l + {3 O}_{2}

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Al and KClO3 react together to form AI2O3 according to 2KClO3⟶2KCl+3O2 4Al+3O2⟶2AI2O34molesof KClO3 (50% pure) on reaction with excess of Al forms x moles of AI2O3.

The correct option is A 2 mol2KClO3⟶2KCl+3O2 4Al+3O2⟶2Al2O350% 4 moles means 2 mole, so, according to balance reaction it can be seen that 2 mole of KClO3 forms 2 moles of Al2O3.

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 $A I_{2} O_{3}$ .

The correct option is A 2 mol
$2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}$
$4 A l + 3 O_{2} ⟶ 2 A l_{2} O_{3}$
50% 4 moles means 2 mole, so, according to balance reaction it can be seen that 2 mole of $K C l O_{3}$ forms 2 moles of $A l_{2} O_{3}$ .