Question

Calculate the volume of hydrogen liberated when 230 grams of sodium reacts with excess of water at STP (Atomic mass of Na = 23, O = 16, H = 1).​

Answer 1

Explanation:

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