Question

Aqueous hydrochloric acid (HCI) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H20). Suppose 1.09 g of hydrochloric acid is mixed with 0.72 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 Hydrochloric acid (HCl)=Molar mass of H+Molar mass of Cl=1 g/mol+35.5 g/mol=36.5 g/mol

Number of moles of HCl=Given mass/Molar mass

=1.09 g/36.5 g/mol=0.03 mol

Molar mass of NaOH=Molar mass of Na+Molar nass of O+Molar mass of H=23 g/mol+16 g/mol+1 g/mol=40 g/mol

Number of moles of NaOH=Given mass/Molar mass

=0.72 g/40 g/mol=0.02 mol

The balanced chemical equation for reaction between HCl and NaOH is as shown

HCl(aq) + NaOH(aq) + NaCl(aq) + H2O(1)

As per the balanced chemical reaction

1 mol NaOH reacts with 1 mol HCl

So 0.02 mol NaOH will react with 0.02 mol HCl

But we have 0.03 mol HCl which is more than the required amount. So HCl is in excess and NaOH is the limiting reagent.

Now as per the balanced chemical reaction

1 mol NaOH will produce 1 mol NaCl

So 0.02 mol NaOH will produce 0.02 mol NaCl

Molar mass of NaCl=Molar mass of Na+Molar mass of Cl

=23 g/mol+35.5 g/mol=58.5 g/mol

So mass of NaCl produced=number of moles of NaCl x Molar mass

=0.02 mol x 58.5 g/mol=1.17 g≈1.2 g (answer rounded to two significant digits as 7 in hundredth place is more than 5 so 1 in tenth place is rounded off to 2)