Question

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 8.40 liters of H2 gas were produced at STP?

Answer 1

Answer : The initial quantity of sodium metal used is 17.25 grams.

Solution : Given,

Volume of $H_2$ gas = 8.40 L

Molar mass of Na metal = 23 g/mole

The Net balanced chemical reaction is,

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

At STP,  22.4 L of volume is occupied by 1 mole of $H_2$ gas

so,  8.40 L of volume is occupied by = $\frac{8.40L\times 1mole}{22.4L}$ = 0.375 moles of $H_2$ gas

Now from the above reaction, we conclude that

1 mole of $H_2$ gas produced by the 2 moles of Na metal

0.375 moles of $H_2$ gas produced = $\frac{2moles\times 0.375mole}{1mole}=0.75moles$ of Na metal

The quantity of Na metal used = Moles of Na metal × Molar mass of Na metal = 0.75 moles × 23 g/mole = 17.25 grams

Therefore, the initial quantity of sodium metal used is 17.25 grams.

Answer 2

Answer:

17.2

Explanation:

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