If the mass of 191 grams nacl reacted with 74 frams of calcium hydroxide and 80 grams of sodium hydroxide is produced, what mass of calcium chloride is produced?

110.98 g/mol

The reaction between NaCl and Ca(OH)₂ is given by the equation;

2NaCl(aq) + Ca(OH)₂(s) → 2NaOH(aq) + CaCl₂(aq)

We are required to determine the mass of CaCl₂ produced,

We will use the following simple steps;

Step 1: Moles of NaCl and Ca(OH)₂ given

Number of moles = Mass ÷ Molar mass

Moles of NaCl

Mass of NaCl = 191 g

Molar mass NaCl = 58.44 g/mol

Number of moles = 191 g ÷ 58.44 g/mol

                             = 3.268 moles

                             = 3.27 Moles

Moles of Ca(OH)₂

Mass of Ca(OH)₂ = 74 g

Molar mass of Ca(OH)₂ = 74.093 g/mol

Number of moles = 74 g ÷ 74.093 g/mol

                             = 0.998 mole

                              = 1.0 mole

However, from the equation  2 moles of NaCl requires 1 mole of Ca(OH)₂

Therefore, from the amount of reactants available NaCl was in excess and Ca(OH)₂ is the limiting reactant .

Step 2: Moles of CaCl₂ produced

From the equation

1 mole of Ca(OH)₂ reacts with NaCl to produce 1 mole of CaCl₂

Therefore; the mole ratio of Ca(OH)₂ to CaCl₂ is 1: 1

Thus;

Moles of CaCl₂ produced is 1.0 moles

Step 3: Mass of CaCl₂ produced

Moles of CaCl₂ = 1.0 mole

Molar mass CaCl₂ = 110.98 g/mol

But; mass = number of moles × Molar mass

Therefore;

Mass of CaCl₂ = 1.0 mole × 110.98 g/mol

                       = 110.98 g CaCl₂