Question

0.52 g of sodium was added to 100 cm3 of water. Calculate: a) The volume of hydrogen evolved at 298 K and 100 kPa. b) The concentration of the sodium hydroxide solution produced, assuming the volume of water does not change. 2Na+ 2H2O + 2NaOH + H2​

Answer 1

Answer:  280 ml and 0.226M NaOH

Explanation:  Three steps:

1)  Calculate the moles of H2 and NaOH produced.  See the graphic.  0.52 grams of Na is 0.0226 moles of Na.  Water is in excess, so assume all the Na is converted to NaOH and H2.  The molar ratio of H2 to Na is 1/2, so we'll only obtain 0.0226/2 = 0.011309 moles of H2 (Blue arrow).  The molar ratio for the NaOH is 1 to 1, so we'll have 0.0226 moles of NaOH.

2)  Use the ideal gas law to find the volume 0.011309 moles of H2 at the given conditions.  PV=nRT.  See the second graphic for the summary of input conditions.  kPa was converted to atm.  I find the volume of H2 produced to be 0.280 liters, or 280 ml.

3)  Find the NaOH concentration.  Despite the caveat that we should assume the volume of water does not change, in fact I find that 0.407 grams is consumed in the reaction.  But, in keeping with the spirit, we'll assume all the original volume, 100 cm^3, is still intact.  At a density of 1 g/cm^3, that is 100 ml of water.  The 0.0226 moles of NaOH is dissolved in 100ml of water.  The concentration, M, is in moles/liter, so to find M:  (0.0226 moles/0.100 liters) = 0.226M NaOH

Wheh . .