Question

2Na + 2H2O → 2NaOH + H2

Answer 1

The initial quantity of sodium metal is 17.25

Answer 2

Answer: 13 grams

Explanation:

$2Na+2H_2O\rightarrow 2NaOH+H_2$

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

According to stoichiometry,

1 mole of $H_2$ is produced from 2 moles of sodium

or 22.4 L of $H_2$ at STP is produced from =2 moles of sodium

thus 6.30 L of $H_2$ at STP is produced from =$\frac{2}{22.4}\times 6.30=0.56moles$ of sodium

To calculate the mass of sodium metal used

$\text{mass of sodium used}=moles\times {\text {Molar mass}}=0.56\times 23=13grams$

Thus the initial quantity of sodium metal used is 13 grams.