Question

Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water. Suppose 7.3 g of hydrochloric acid is mixed with 11.7 g of sodium hydroxide. Calculate the maximum mass of sodium chloride that could be produced by the chemical reaction. Round your answer to 2 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)= 7.3 g

number of mol of HCl,
n = mass of HCl/molar mass of HCl
=(7.3 g)/(36.458 g/mol)
= 0.2002 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)= 11.7 g

number of mol of NaOH,
n = mass of NaOH/molar mass of NaOH
=(11.7 g)/(39.998 g/mol)
= 0.2925 mol
Balanced chemical equation is:
HCl + NaOH ---> NaCl + H2O


1 mol of HCl reacts with 1 mol of NaOH
for 0.2002 mol of HCl, 0.2002 mol of NaOH is required
But we have 0.2925 mol of NaOH

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


Molar mass of NaCl,
MM = 1*MM(Na) + 1*MM(Cl)
= 1*22.99 + 1*35.45
= 58.44 g/mol

According to balanced equation
mol of NaCl formed = (1/1)* moles of HCl
= (1/1)*0.2002
= 0.2002 mol


mass of NaCl = number of mol * molar mass
= 0.2002*58.44
= 11.7 g

Answer: 12 g