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Standard XII

Chemistry

Nitrogen

For the react...

Question

For the reaction
$A l + O_{2} \to A l_{2} O_{3}$
(The equation is not balanced)
54 g of Al is reacted with 67.4 litre at STP. 76.5 g of $A l_{2} O_{3}$ was obtained. The percentage yield of the product obtained is

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Solution

The balanced chemical equation is
$4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$

108 g of $A l$ on reaction with 67.4 litre of oxygen at STP gives 204 g of alumina.
54 g of $A l$ should react with 33.7 litre of oxygen and give 102 g of alumina.

Hence the amount of excess reagent = 67.4 - 33.7 = 33.7 litre
Theoretical yield = 102 g
Experimental yield = 76.5 g
Percentage yield $= \frac{experimental yield}{theoretical yield} \times 100$
$= \frac{76.5}{102} \times 100$
$= 75 %$

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Similar questions

For the reaction
$A l + O_{2} \to A l_{2} O_{3}$
$54 g$ of $A l$ is reacted with $67.4$ litre of $O_{2}$ at STP. $76.5 g$ of $A l_{2} O_{3}$ was obtained. The percentage yield of the product obtained is

27 g m

A l

react with an excess of oxygen to give

4.59 g m

A l_{2} O_{3}

. Calculate the percentage yield of the reaction.

Q. The mass of

A l (O H)_{3}

that was obtained by the reaction of 133.4 g of

A l C l_{3}

with 40 g of

N a O H

according to following equation is 13 g.

A l C l_{3} + N a O H \to A l (O H)_{3} + N a C l

(not balanced)
Calculate the percentage yield.

(Molar mass of aluminium chloride is 133.34 g/mol, molar mass of

N a O H

is 40 g/mol and molar mass of aluminium hydroxde is 78 g/ mol respectively)

S_{2} O_{8}^{2 -} + I^{-} S O_{4}^{2 -} + I_{2}

What is the maximum yield of iodine that can be obtained when 1 mole of

N a_{2} S_{2} O_{8}

reacts completely with excess iodide ion according to the equation above? (Equation above is not balanced.)

Q. In a certain reaction 389.5 g of

T i C l_{4}

is reacted with 96 g of

M g

. Calculate the percentage yield of

T i

if 32 g of

T i

is actually obtained .

[A t . w t T i = 48 u, M g = 24 u C l = 35.5 u]

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For the reaction Al+O2→Al2O3 (The equation is not balanced) 54 g of Al is reacted with 67.4 litre at STP. 76.5 g of Al2O3 was obtained. The percentage yield of the product obtained is

For the reaction
$A l + O_{2} \to A l_{2} O_{3}$
(The equation is not balanced)
54 g of Al is reacted with 67.4 litre at STP. 76.5 g of $A l_{2} O_{3}$ was obtained. The percentage yield of the product obtained is