Question

Aqueous hydrochloric acid (HCl) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H2O). Suppose 32.5 g of hydrochloric acid is mixed with 19. g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Molar mass of HCl,
MM = 1*MM(H) + 1*MM(Cl)
= 1*1.008 + 1*35.45
= 36.458 g/mol


mass(HCl)= 32.5 g

number of mol of HCl,
n = mass of HCl/molar mass of HCl
=(32.5 g)/(36.458 g/mol)
= 0.8914 mol

Molar mass of NaOH,
MM = 1*MM(Na) + 1*MM(O) + 1*MM(H)
= 1*22.99 + 1*16.0 + 1*1.008
= 39.998 g/mol


mass(NaOH)= 19.0 g

number of mol of NaOH,
n = mass of NaOH/molar mass of NaOH
=(19.0 g)/(39.998 g/mol)
= 0.475 mol
Balanced chemical equation is:
HCl + NaOH ---> H2O + NaCl


1 mol of HCl reacts with 1 mol of NaOH
for 0.8914 mol of HCl, 0.8914 mol of NaOH is required
But we have 0.475 mol of NaOH

so, NaOH is limiting reagent
we will use NaOH in further calculation


Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol

According to balanced equation
mol of H2O formed = (1/1)* moles of NaOH
= (1/1)*0.475
= 0.475 mol


mass of H2O = number of mol * molar mass
= 0.475*18.02
= 8.6 g

Answer: 8.6 g