Question

If 11.2 kilograms of Al2O3(s), 52.4 kilograms of NaOH(l), and 52.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?What is the total mass of the excess reactants left over after the reaction is complete?

Answer 1

first write the balance dequation

Al₂O₃ + 6 NaOH + 12 HF ------> 2 Na₃AlF6 + 9 H2O

no of moles of Al2O3 = 11200 g / 101.96 g/mol = 109.85 moles

no fo moles of NaOH = 52400 / 40 = 1310 moles

no o fmoles of HF = 52400 g / 20 = 2620 moles

from the balanced equation to get 2 moles of cryolite we need one mole of Al2O3 6 moles of NaOH and 12 moles of HF

but here limiting agent is Al2O3

from the balanced equation it is cler that 1 mole of Al2O3 will give 2 moles of Cryolite

109.85 moles moles of Al2O3 will give 2 x 109.85 = 219.7 moles of cyolite should get

now we know the moles of cryolite

weight of cryolite = moles x molar mass of cryolite

= 219.7 moles x  x 209.94 g/mole

= 46123.82 grams or

= 46.123 Kg cryolite

for 1 mole of Al2O3 6 moles of NaOH required

no of moles of NaOH remain = 1013 - 659.1 =353.9 moles of NaOH left

for one mole of Al2O3 required 12 moles of Hf

no of moles of HF remaining = 2620 - 1318.2 = 1301.8 moles of Hf remaining