Question

Cryolite, Na, AIF,(), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation for the synthesis of cryolite. equation: A1,0,($) +6 NaOH(1) + 12 HF(g) 2 Na, AIF+9H2O(g) If 13.4 kg of Al,0,(s), 51.4 kg of NaOH(I), and 51.4 kg of HF(g) react completely, how many kilograms of cryolite will be produced? mass of cryolite produced: 55 kg Na, AlF Which reactants will be in excess? HF A1,03 NaOH What is the total mass of the excess reactants left over after the reaction is complete? total mass of excess reactants:

How do we find the total mass of excess reactants?

Answer 1

Solution:

For the total excess mass of reactant, we first calculate tge number of moles of all reactants.

Number of moles of all the reactants are calculated as,

>Number of moles of Al2O3 = Mass / molar mass

= 13.4 kg / 101.96 g mol-1 = 1 3400 g / 101.96 g mol-1

= 131.424 mol

>Number of moles of NaOH = mass /molar mass

= 51.4 kg / 40 g mol-1 = 51400 g / 40 g mol-1

= 1285 mol

> Number of moles of HF = mass / molar mass

= 51.4 kg / 20.01 g mol-1 = 51400 g / 20.01 g mol-1

= 2568.72 mol

From the equation for the formation of cryolite, it can be infered that, 1 mol Al2O3 requires 6 mol of NaOH and 12 mol of HF.

Thus,

Required moles of NaOH = 6 x number of mol of Al2O3

= 6 x 131.424 mol = 788.544 mol

Thus, excess mol of NaOH = 1285 - 788.544 = 496.46 mol

Required moles of HF = 12 x number of mol of Al2O3

= 12 x 131.424 mol = 1577.088 mol

Excess mol of HF = 2568.72 - 1577.088 = 991.632 mol

>>Thus, from calculation, it can be seen that max number of moles which is in excess belongs to HF.

>> Mass of excess reactant (HF)

= excess number of moles x molar mass

= 991.632 mol x 20.01 g mol-1 = 19882.56 g

= 19.883 kg

>> mass of excess reactant (NaOH)

= excess moles x molar mass

= 496.46 mol x 40 g mol-1 = 19859.4 g

= 19.859 kg

Hence,

Total excess mass = excess mass of NaOH + excess mass HF

= 19.859 kg + 19.883 kg = 39.741 kg