Question

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?

Answer 1

Answer:

110.98 g/mol

Explanation:

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