A student is working on a limiting reactant and percent yield problem, their work is shown below. They find out that the answer they wrote is incorrect, look over the work and find/highlight and explain the mistakes made; use a different colour for your notes. 13.5g H2SO4 reacts with 250mL of 0.76 M solution of NaOH, find the limiting reactant and percent yield of sodium sulfate if 10.5g is produced.

H2SO4 + 2 NaOH → 2 H2O + Na2SO4

13.5g H2SO4
98.09 mol H2SO4
1 mol Na2SO4

142.05 g Na2SO4

1 g H2SO4
1 mol H2SO4
1 mol Na2SO4

= 18,800 g Na2SO4
This is the limiting reactant

0.76 M = x / 250 = 190 mol

190 mol NaOH
1 mol Na2SO4
142.05 g Na2SO4

1 mol NaOH
1 mol Na2SO4

= 27, 000 g Na2SO4
This is the excess reactant

10.5 / 18800 * 100 = 0.056 %