Question #54740
2 Answers
This answer will assume the hydrogen is formed at STP (allows us to use molar volume of a gas of 22.4L). Standard pressure is actually 0.987atm (100kPa).
2.5g Na x (1 mol Na/22.99g Na) = 0.1087 mol Na
*convert to moles of sodium
0.1087 mol Na x (1 mol H2/2 mol Na) = 0.05437 mol H2
*convert from moles sodium to moles Na using ratio from balanced eq
0.05437 mol H2 x (22.4 L H2 / 1 mol H2) = 1.218 L H2
*convert from mol hydrogen to volume hydrogen using molar volume
The answer should be rounded to 2 sig figs based on the mass measurement of 2.5g Na you started with so your final answer is
1.2L Hydrogen at 1 atm
1.2 L.
Explanation:
The volume of H₂ is 1.2 L.
2Na + 2H₂O → 2Na⁺ + 2OH⁻ + H₂
You must convert:
grams of Na → moles of Na → moles of H₂ → litres of H₂

2.5 g Na ×
#(1 mol Na)/(22.99 g Na)# = 0.11 mol Na 
0.11 mol Na × #(1 mol H₂)/(2 mol Na)* = 0.054 mol H₂

At 1 atm and 0 °C, 1 mol of H₂ has a volume of 22.414 L.
0.054 mol H₂ ×