Question

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation.

Al2O3(s) + NaOH(l) + HF(g) ---> Na3AlF6 + H2O(g)

If 11.3 kilograms of Al2O3(s), 60.4 kilograms of NaOH(l), and 60.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?

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(l) + 12HF(g) ---> 2Na3AlF6 + 9H2O(g)

m = 11.3 kg = 11300 g of Al2O3

m = 60.4 = 60400 g of NaOH

m = 60.4 = 60400 g of HF

find cryolite formed

MW of Al2O3 = 101.96

MW of NaOH = 40

MW of HF = 20.01

mol of Al2O3 = mass/MW = 11300 /101.96 = 110.827

mol of NaOH = mass/MW = 60400 /40 = 1510

mol of HF = mass/MW = 60400 /20.01 = 3018.490

recall ratios:

Al2O3(s) + 6NaOH(l) + 12HF(g)

fin limiting reactant

110.827 need 6*110.827 = 664.962 mol of NaO and 110.827 *12 = 1329.924 mol of HF

clearly, Al2O3 is limiting reactant

then

110.827 mol of Al2O3 will form 2*110.827= 221.654 mol of Na3AlF6

mass = mol*MW = 221.654 *209.94 = 46534.04076 mass = 46.53 kg of Cryolite

as stated before

Al2O3 is limiting the reactiong

total mass excess:

mass of NaOH + mass of HF

mol of NaOH left = 1510 - 664.962 = 845.038 mol of NaOH left

mass of NaOH = mol*MW = 845.038 *40 = 33801.52

mol of HFleft = 3018.490 - 1329.924 = 1688.566 mol of HFleft

mass HF = mol*MW = 1688.566 *20 = 33771.32

total mass = 33801.52+33771.32 = 67572.84 g = 67.5724 kg left of excess reactants