Question

23.balance Al2O3(s)+NaOH(i)+Hf(g)--> Na3AlF6+H2O(g)

a. If 11.1 kilograms of Al2O3(s), 59.4 kilograms of NaOH(l), and 59.4 kilograms of HF(g) react completely, how b. many kilograms of cryolite will be produced?

c. Which reactants will be in excess?

What is the total mass of the excess reactants left over after the reaction is complete?

Answer 1

Al₂O₃(s) + 6NaOH(aq) + 12HF(g) -------> 2Na₃AlF₆ + 9H₂O(g)

Molar mass of Al₂O₃ = 102 g/mole

Molar mass of NaOH = 40 g/mole

Molar mass of HF = 20 g/mole

Now, moles of Al₂O₃ in 11100 g of it = mass/molar mass = 108.824

Moles of NaOH in 59400 g of it = mass/molar mass = 59400/40 = 1485

Moles of HF in 59400 g of it = = mass/molar mass = 2970

Now, as per the balanced reaction:-

Al₂O₃, NaOH & HF reacts in the molar ratio of 1:6:12

Thus, for 108.824 moles of Al₂O₃ for complete reaction, 652.941 moles of NaOH & 1305.88 moles of HF are required.

Clearly, NaOH and HF are in excess

Thus, excess moles of NaOH = 1485-652.941 = 832.049

mass of excess NaOH = moles*molar mass = 832.049*40 = 33282.36 g = 33.282 kg

excess moles of HF = 2970 - 1305.88 = 1664.12

Thus, mass of HF left unreacted = moles*molar mass = 1664.12*20 = 33282.4 g = 33.282 kg

Now, moles of Na₃AlF₆ formed = 2*moles of Al₂O₃ reacted = 2*108.824 = 217.648

Molar mass of Na₃AlF₆ = 216 g/mole

Hence, mass of f Na₃AlF₆ formed = moles*molar mass = 217.648*216 = 47011.968 g = 47.012 kg