In: Chemistry
If 11.6 kilograms of Al2O3(s), 53.4 kilograms of NaOH(l), and 53.4 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? Which reactants will be in excess?
Molar mass of Al2O3 is 102 g/mol
Molar mass of NaOH is 40 g/mol
Molar mass of HF is 20 g/mol
Molar mass of cryolite, Na3AlF6 is 210 g/mol
The balanced molecular equation is Al2O3(s) + 6 NaOH(l) + 12 HF(g) ----> 2Na3AlF6 + 9H2O
According to the balanced reaction ,
1 mol=102 g of Al2O3 reacts with 6 mol=6x40=240 g of NaOH & 12mol=12x20=240 g of HF
OR
102 kg of Al2O3 reacts with 240 kg of NaOH & 240 kg of HF
11.6 kg of Al2O3 reacts with M kg of NaOH & N kg of HF
M = ( 11.6x240) / 102 & N = (11.6x240) /102
= 27.3 kg = 27.3 kg
So 53.4-27.3 = 26.1 kg of NaOH left unreacted
And 53.4 - 27.3 = 26.1 kg of HF also left unreacted
Therefore NaOH & HF are the excess reactants.
From the balanced equation,
1 mole =102 g of Al2O3 produces 2 mol = 2x210 =420 g of cryolite
OR
102 kg of Al2O3 produces 420 kg of cryolite
11.6 kg of Al2O3 produces Y kg of cryolite
Y = ( 11.6x420) / 102
= 47.8 kg of cryolite
Therefore the mass of cryolite produced is 47.8 kg