In: Chemistry
What is the limiting reactant if 2.047 grams of NaOH (s) reacts with 101.3 mL of a 2.09 M HCl solution?
Molar mass of NaOH = At.mass of Na + At.mass of O + At.mass of H
= 23+16+1
= 40 g/mol
Given mass of NaOH , m = 2.047 g
So number of moles of NaOH , n = mass/molar mass
= 2.047 / 40
= 0.0512 moles
Number of moles of HCl , n' = Molarity x volume in L
= 2.09 M x 0.1013 L
= 0.212 moles
The balanced equation is : HCl + NaOH ---> NaCl + H2O
According to the balanced equation ,
1 mole of NaOH reacts with 1 mole of HCl
0.0512 mole of NaOH reacts with 0.0512 moles of HCl
So (0.212 - 0.0512 ) moles of HCl left unreacted so HCl is the excess reactant.
Since all the mass of NaOH completly reacted , NaOH is the limiting reactant.