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?

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

According to stoichiometry,

1 mole of [tex]H_2[/tex] is produced from 2 moles of sodium

or 22.4 L of [tex]H_2[/tex] at STP is produced from =2 moles of sodium

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

To calculate the mass of sodium metal used

[tex]\text{mass of sodium used}=moles\times {\text {Molar mass}}=0.56\times 23=13grams[/tex]

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

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?

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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?