If 3.0x10^23 atoms of sodium and 250 mL of 3.0 M HCl solution react to give 4.9L of a gas at 25C and 760 torr, what is the percent yield of that gas?

Answer 1

Write a correctly balanced equation for the reaction taking place:

2Na(s) + 2HCl ==> 2NaCl(s) + H2(g)

moles Na present = 3.0x10²³ atoms Na x 1 mol Na/6.02x10²³ atoms = 0.50 moles Na

moles HCl present = 250 ml x 1 L/1000 ml x 3.0 mol/L = 0.75 moles HCl

Which reactant is limiting? In the balanced equation the mol ratio is 1:1, so in this example Na is limiting.

Theoretical yield of H2(g) based on 0.50 moles Na:

0.50 mol Na x 1 mol H2/2 mol Na = 0.25 moles H2

Using PV = nRT and solving for the volume (V) of H2, we have...

V = nRT/P = (0.25 mol)(62.36 L-torr/Kmol)(298K)/760 torr = 6.11 L = theoretical yield of H2(g)

% yield = actual yield/theoretical yield (x100%) = 4.9 L / 6.11 L (x100%) = 80%