Cryolite, Na3AlF6 (s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation. 1.) Balance the equation - AlO3 (s) + NaOH (l) + HF (g) - ->Na3AlF6+H2O (g) 2.) If 17.5 kilograms of Al2O3 (s), 51.4 kilograms of NaOH (l), and 51.4 kilograms of HF (g) react completely, how many kilograms of cryolite will be produced?3.) Which reactants will be in excess, (Al2O3, NaOH, or HF) 4.) What is the total mass of the excess reactants left over after the reaction is complete in KG?

(1) Al2O3 + 6NaOH + 12HF = 2Na3AlF6 + 9H2O

(2) 72.1 kilograms of cryolite would be produced

(3) The excess reactants would be NaOH and HF

(4) The total mass of reactants left over after the reaction is complete is 20.4 kg

Explanation:

(1) One mole of aluminum oxide will react with six moles of sodium hydroxide and twelve moles of hydrogen fluoride to produce two moles of cryolite and nine moles of water.

(2) From the equation of reaction:

1 kgmole of Al2O3 (102kg) produced 2 kgmoles of Na3AlF6 (420 kg)

17.5 kg of Al2O3 would produce 17.5*420/102 = 72.1 kg of cryolite

(2) 1 kgmole of Al2O3 (102kg) reacted with 6 kgmoles of NaOH (240kg)

17.5 kg of Al2O3 would reat with 17.5*240/102 = 41.2 kg of NaOH

41.2 kg of NaOH is required to react completely with 17.5 kg of Al2O3 but 51.4 kg of NaOH is present, so NaOH is an excess reactant

1 kgmole of Al2O3 (102kg) reacted with 12 kgmoles of HF (240 kg)

17.5 kg of Al2O3 would react with 17.5*240/102 = 41.2 kg of HF

HF is also an excess reactant because 41.2 kg is needed to completely react with 17.5 kg of Al2O3 but 51.4 kg is present

Therefore, the excess reactants are NaOH and HF

(4) Total mass of excess reactants = (51.4 - 41.2) + (51.4 - 41.2) = 10.2 + 10.2 = 20.4 kg

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

2. 72.1 kg.

3. NaOH and HF

4. 20.45 kg

Explanation:

1.

Equation of the reaction

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

2.

Molar masses of reactants

Al2O3

= (27*2) + (16*3)

= 102 g/mol.

NaOH

= 23 + 16 + 1

= 40 g/mol.

HF

= 1 + 19

= 20 g/mol

Moles of reactants

Number of moles = mass/molar mass

Al2O3

= 17500/102

= 171.57 moles

NaOH

= 51400/40

= 1285 moles

HF

= 51400/20

= 2570 moles

Using stoichiometry, from the balanced equation above;

1 mole of Al2O3 reacts with 6 moles of NaOH and 12 moles of HF.

Calculating how many moles of HF will be required if:

• NaOH is used up.

= 2570 moles of HF * (6 moles of NaOH) / 12 moles of HF

= 1285 moles of NaOH will be required.

• Al2O3 is used up.

= 2570 moles of HF * (1 moles of Al2O3) / 12 moles of HF

= 214.17 moles of Al2O3 will be required.

But there are only 171.57 moles of Al2O3. Therefore, Al2O3 is the limiting reagent.

Therefore, since 2 moles of cryolite was produced from 1 mole of Al2O3.

There are 171.56 * 2

= 343.12 moles of cryolite.

Molar mass of cryolite, Na3AlF6 =

(23*3) + 27 + (19*6)

= 210 g/mol.

Mass of Na3AlF6 = mass*molar mass

= 210 * 343.12

= 72,059.4 g

= 72.1 kg.

3.

To calculate excess reactants:

Since the limiting reagent is Al2O3 with 171.57 moles.

Therefore, for;

NaOH

= 6*171.56

= 1029.42 moles of NaOH reacted.

Excess number of moles = moles of supplied NaOH - moles of supplied NaOH

= 1285 - 1029.42

= 255.58 moles of NaOH is in excess.

HF

= 12*171.57

= 2058.84 moles of HF reacted.

Excess number of moles = moles of supplied HF - moles of supplied HF

= 2570 - 2058.84

= 511.16 moles of HF is in excess.

Therefore, NaOH and HF are reactants in excess.

4.

Excess masses

NaOH

= 40*255.58

= 10223.3 g

= 10.223 kg

HF

= 20*511.16

= 10223.2 g

= 10.223 kg

Total masses in excess = 10.223 + 10.223

= 20.45 g of excess reactants.