Question

NaOH is formed according to reaction,
$2\text{Na}+\dfrac{1}{2}{{\text{O}}_{2}}\to \text{Na}{{\text{O}}_{2}}$
$\text{Na2 O}+{{\text{H}}_{2}}\text{O}\to 2\text{NaOH}$
To make 4 g NaOH, N a required is:

Answer 1

Hint Check the given reaction and find the no. of moles of sodium used in the formation of the final sodium hydroxide. We can see in the above given reaction that 2 moles of sodium hydroxide are formed as end product which are formed from $N{{a}_{2}}O$ and water i.e. 2 moles of Na are being used in order to make 2 moles of sodium hydroxide. So, simply find the no. of moles in 4g of $NaOH$ and then find the mass for the same amount of moles for Na.
Formula Used: No. of moles = given mass/molar mass

Complete Step By Step Solution:
2 moles of $NaOH$ are being formed from 2 moles of Na. The molar mass of $\text{NaOH}=23+16+1$
= 40
So, no. of moles in 4g of $\text{NaOH}=\dfrac{4}{40}=0\cdot 1$
Since the same moles will be required for Na so we will find the mass of 0.1 mole of Na.
The molar mass of Na = 23g per mole (obtained by multiplying the atomic no. by 2 and then adding 1 because atomic no. is odd no. The atomic no. of sodium is 11, if atomic no. of an element is even no. then simply multiply by 2 to get the molar mass of that element)
Mass of 0.1 mole of Na $=23\times 0\cdot 1$
$=2\cdot 3g$.

Hence the correct choice is (A).

Additional Information:
Sodium hydroxide is an organic compound which is also known as caustic soda. It is mostly used in metal processing, chemical manufacturing and for water treatment processes. It is also used as a household bleaching agent. It can cause eye and skin irritation. Sodium hydroxide is soluble in water, ethanol and methanol.

Note: Molar mass or atomic mass for an element is obtained by multiplying its atomic no. by 2 and adding 1 to it depending on its atomic no. being odd or even whereas molar mass for a compound is obtained by adding the atomic masses of its constituent elements like for sodium hydroxide the constituent elements are sodium, oxygen and hydrogen.