Question

If 12.2 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produced?

Answer 1

Answer:  kilograms of cryolite ≈ 50.23 kg

Explanation:

Write the balanced equation

Let's start off by writing the reaction for the formation of cryolite:

Al₂O₃  +  6NaOH  +  12HF    →    2Na₃AlF₆  +  9H₂O

Determine the limiting factor

moles = mass ÷ molar mass

moles of Al₂O₃ = 12200 g ÷ 101.96 g/mol = 119.65 mol

moles of NaOH = 57400 g ÷  39.997 g/mol = 1435.11 mol

moles of HF = 57400 g ÷ 20.01 g/mol = 2868.57 mol

Since Al₂O₃ has the smallest number of moles in the reaction, it is the limiting factor. As such we will have to use the moles of Al₂O₃ to determine the moles of cryolite that will be produced.

Find the mole ratio between the limiting factor and the cryolite

The mole ratio of Al₂O₃ : Na₃AlF₆ based on the balanced equation is 1 : 2

∴ since the moles of Al₂O₃ = 119.65 mol

        the moles of Na₃AlF₆ = 119.65 mol × 2 = 239.3 mol

Calculate the mass of the Cryolite

Mass cryolite = moles × molar mass

                      = 239.3 mol × 209.9 g/mol

                      =  50229.07 g

      ∴ kilograms of cryolite ≈ 50.23 kg