Question

Given :

NaOH (aq) + HCl (aq) -> H2O (l) + NaCl (aq)

Mass of 1.00M HCl (diluted in water) : 60.925g

Vol of 1.00M HCl : 60.3mL

Mass of 1.00M NaOH : 50.500g

Vol of 1.00M NaOH : 50.0mL

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1. Find mole of product if: (show calculations)

A. HCl is the limiting reactant: ______

B. NaOH is the limiting reactant: ______

2. FInd the actual limiting reactant: ______

Answer 1

1) no. of mole = Molarity volume of solution in liter

60.3 ml = 0.0603 liter

no. of mole of HCl = 1.00 0.0603 = 0.0603 mole

50 ml = 0.050 liter

no. of mole of NaOH = 1.00 0.050 = 0.050 mole

According to reaction 1 mole of HCl react with 1 mole of NaOH then to react with 0.0603 mole of HCl required 0.0603 mole of NaOH but NaOH given only 0.050 mole therefore NaOH is limiting reactant react completly.

According to 1 mole of each produce 1 mole of NaCl and 1 mole of water then 0.050 mole of NaOH and 0.050 mole of HCl produce 0.050 mole of NaCl and 0.050 mole of H₂O

2)

no. of mole = Molarity volume of solution in liter

60.3 ml = 0.0603 liter

no. of mole of HCl = 1.00 0.0603 = 0.0603 mole

50 ml = 0.050 liter

no. of mole of NaOH = 1.00 0.050 = 0.050 mole

According to reaction 1 mole of HCl react with 1 mole of NaOH then to react with 0.0603 mole of HCl required 0.0603 mole of NaOH but NaOH given only 0.050 mole therefore NaOH is limiting reactant