<h2>
Answer: kilograms of cryolite ≈ 50.23 kg</h2>
Explanation:
<u><em>Write the balanced equation</em></u>
Let's start off by writing the reaction for the formation of cryolite:
Al₂O₃ + 6NaOH + 12HF → 2Na₃AlF₆ + 9H₂O
<em><u>Determine the limiting factor</u></em>
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.
<em><u>Find the mole ratio between the limiting factor and the cryolite</u></em>
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
<u><em>Calculate the mass of the Cryolite</em></u>
Mass cryolite = moles × molar mass
= 239.3 mol × 209.9 g/mol
= 50229.07 g
∴ kilograms of cryolite ≈ 50.23 kg
