determine the mol ratio of both the reagents Then calculate the mol for both reagents eg NaOH + HCl --> NaCl + H2O given in qns NaOH is 5g and 50cm^3 of 0.2mol/dm^3 HCL
from eqn: Mole ratio of NaOH to HCL = 1:1
from calculations: no. of mole of NaOH = 5 / (23+16 +1) = 0.125mol (excess) no. of mole of HCL = 0.2 * 50/1000 =0.01 mol (limiting)
NaOH : HCL should equals to 1:1 however from calculations it is not 1:1 and n(HCL) < n(NaOH) HCL limiting, NaOH excess
