What is the limiting reactant if 2.047 grams of NaOH (s) reacts with 101.3 mL of...

What is the limiting reactant if 2.047 grams of NaOH (s) reacts with 101.3 mL of a 2.09 M HCl solution?

Expert 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.

queen_honey_blossom answered 6 months ago

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