Question

Read the given equation.

2Na + 2H2O → 2NaOH + H2

During a laboratory experiment, a certain quantity of sodium metal reacted with water to produce sodium hydroxide and hydrogen gas. What was the initial quantity of sodium metal used if 6.30 liters of H2 gas were produced at STP?

a. 10.3 grams
b. 12.9 grams
c. 14.7 grams
d.15.2 grams

Answer 1

Answer: b. 12.9 grams

Explanation:

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

According to avogadro's law, 1 mole of every gas occupies 22.4 Liters at STP ,contains avogadro's number
`6.023* 10^(23)` of particles and weighs equal to the molecular mass.

Thus at STP :

1 mole of hydrogen is produced by 2 moles of sodium metal

i.e. 22.4 L of
`H_2` is produced by 46 gram of sodium metal

6.3 L of
`H_2` is produced by=
`(46)/(22.4)* 6.3=12.9gram` of sodium metal