Question

Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H2O . Suppose 27. g of hydrochloric acid is mixed with 11.2 g of sodium hydroxide. Calculate the minimum mass of hydrochloric acid that could be left over 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)= 27.0 g

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(27.0 g)/(36.458 g/mol)

= 0.7406 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)= 11.2 g

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(11.2 g)/(39.998 g/mol)

= 0.28 mol

Balanced chemical equation is:

HCl + NaOH ---> NaCl + H2O

1 mol of HCl reacts with 1 mol of NaOH

for 0.740578 mol of HCl, 0.740578 mol of NaOH is required

But we have 0.280014 mol of NaOH

so, NaOH is limiting reagent

we will use NaOH in further calculation

According to balanced equation

mol of HCl reacted = (1/1)* moles of NaOH

= (1/1)*0.280014

= 0.280014 mol

mol of HCl remaining = mol initially present - mol reacted

mol of HCl remaining = 0.740578 - 0.280014

mol of HCl remaining = 0.4606 mol

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass of HCl,

m = number of mol * molar mass

= 0.4606 mol * 36.458 g/mol

= 17 g

Answer: 17 g