
Complete the following reaction: $2NaOH + 2Al \to $
A. $2NaAl{O_2} + {H_2}$
B. $2NaAl{O_2} + \frac{1}{2}{H_2}$
C. $3NaAl{O_2} + 3{H_2}$
D. $2NaAl{O_2} + 2{H_2}$
Answer
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Hint:When an alkali or base reacts with metal, it will produce salt and hydrogen gas is evolved as shown
Alkali + Metal $\to$ Salt + Hydrogen
Examples are sodium hydroxide gives hydrogen gas and sodium zincate reacts with zinc metal.
Complete answer:
As metal reacts with base it will evolve hydrogen gas and a salt has formed. In case of aluminium reaction with base that is sodium hydroxide will produce sodium aluminate as salt and hydrogen gas.
The reaction between sodium hydroxide and aluminium can be written as
$NaOH + Al \to NaAl{O_2} + {H_2}$
This is not a balanced reaction. So we have to balance the above reaction.
In the reactant side 1 hydrogen is present but product side 2 hydrogen are there. So to balance hydrogen on both sides, multiply $NaOH$on the reactant side by 2. Now the reaction will become-
$2NaOH + 2Al + 2{H_2}O \to 2NaAl{O_2} + 3{H_2}$
$2NaOH + Al \to NaAl{O_2} + {H_2}$
Now hydrogen is balanced on both the side, but sodium is not balanced so multiply $NaAl{O_2}$ by 2 to balance out. Now our reaction will become-
$2NaOH + Al \to 2NaAl{O_2} + {H_2}$
Sodium is now balanced. Now the aluminium on the reactant side is 1 but on the product side it is 2. So we will multiply $Al$ by 2. And our reaction has now become-
$2NaOH + 2Al \to 2NaAl{O_2} + {H_2}$
Now hydrogen is balanced, sodium is also balanced and aluminium also balanced. Now we will check for oxygen, reactant side there are 2 oxygen and in product side but 4 oxygen on product side .
To balance out the hydrogen and oxygen we have to add ${H_2}O$ on the left side. And then balance the sodium, hydrogen, oxygen and aluminium. The balanced reaction we will get is
$2NaOH + 2Al + 2{H_2}O \to 2NaAl{O_2} + 3{H_2}$
So the correct option will be $2NaAl{O_2} + 3{H_2}$
Note:
When acid reacts with metals then also salt form and hydrogen gas evolves. The metal present will displace hydrogen from the acid, and that can be seen as hydrogen gas. Then the metal combines with the rest part of the acid and forms a compound called salt.
Alkali + Metal $\to$ Salt + Hydrogen
Examples are sodium hydroxide gives hydrogen gas and sodium zincate reacts with zinc metal.
Complete answer:
As metal reacts with base it will evolve hydrogen gas and a salt has formed. In case of aluminium reaction with base that is sodium hydroxide will produce sodium aluminate as salt and hydrogen gas.
The reaction between sodium hydroxide and aluminium can be written as
$NaOH + Al \to NaAl{O_2} + {H_2}$
This is not a balanced reaction. So we have to balance the above reaction.
In the reactant side 1 hydrogen is present but product side 2 hydrogen are there. So to balance hydrogen on both sides, multiply $NaOH$on the reactant side by 2. Now the reaction will become-
$2NaOH + 2Al + 2{H_2}O \to 2NaAl{O_2} + 3{H_2}$
$2NaOH + Al \to NaAl{O_2} + {H_2}$
Now hydrogen is balanced on both the side, but sodium is not balanced so multiply $NaAl{O_2}$ by 2 to balance out. Now our reaction will become-
$2NaOH + Al \to 2NaAl{O_2} + {H_2}$
Sodium is now balanced. Now the aluminium on the reactant side is 1 but on the product side it is 2. So we will multiply $Al$ by 2. And our reaction has now become-
$2NaOH + 2Al \to 2NaAl{O_2} + {H_2}$
Now hydrogen is balanced, sodium is also balanced and aluminium also balanced. Now we will check for oxygen, reactant side there are 2 oxygen and in product side but 4 oxygen on product side .
To balance out the hydrogen and oxygen we have to add ${H_2}O$ on the left side. And then balance the sodium, hydrogen, oxygen and aluminium. The balanced reaction we will get is
$2NaOH + 2Al + 2{H_2}O \to 2NaAl{O_2} + 3{H_2}$
So the correct option will be $2NaAl{O_2} + 3{H_2}$
Note:
When acid reacts with metals then also salt form and hydrogen gas evolves. The metal present will displace hydrogen from the acid, and that can be seen as hydrogen gas. Then the metal combines with the rest part of the acid and forms a compound called salt.
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