Question

Calculate the volume mass and number of molecules of hydrogen liberated when 230 g of sodium reacts with excess of water of STP (atomic mass of Na equal to 230, o is equal to 16u and H is equal to 1 u

Answer 1

Given that,

Atomic masses of sodium =23u

Oxygen =16u

Hydrogen =1u  

Now,

 2Na  

(g)

​

+2H  

2

​

O→2NaOH  

(aq)

​

+H  

2

​

(g)↑

(2×23)u+2(2×1+1×16)u→2(23+16+1)u+(2×1)u

46u+36u→80u+2u

46g+36g→80g+2g

As per the balanced equation

46g of Na = 2g of Hydrogen

230 g of Na =230×  

46

2g

​

=10g of hydrogen

I gram molar mass of any gas at STP

22.4 liters know as gram molar volume

2.0 g of hydrogen occupies =10.0g×  

2.0g

22.4

​

=112 liters

2 g of hydrogen, 1 mole of H  

2

​

 contains 6.02×10  

23

 x H molecules 10g of hydrogen contain

 =  

2.0g

10.0g×6.02×10  

23

 

​

 

=30.10×10  

23

 

=3.01×10  

24

molecules

Hence, this is the required solution