HCl (aq) + NaOH (aq) →NaCl (aq) + H2O (l)

A student was given the task of titrating a 20. mL sample of 0.10MHCl (aq) with 0.10MNaOH (aq). The HCl (aq) was placed in an Erlenmeyer flask. An equation for the reaction that occurs during the titration is given above.

f) If the student started with 46 grams of NaOH, how much H2O should be produced?

0.036 g

Explanation:

0.036 g of water should be produced.

From the equation of reaction, 1 mole of HCl requires 1 mole of NaOH in order to produce 1 mole of H2O.

20 mL of 0.10 M HCl contains 20/1000 x 0.10 = 0.002 moles of HCl

46 grams of NaOH contains 46/40 = 1.15 moles of NaOH

It thus means that the HCl is a limiting reagent in the reaction.

From the equation:

1 mole HCl will produce 1 mole of H2O.

0.002 HCl with therefore produce 0.002 x 1/1 = 0.002 mole of H2O

Mass of water produced = mole x molar mass

0.002 x 18 = 0.036 g.

Hence, 0.036 g of water would be produced.