Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesi

Question

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

Question 18 of 20 Map University Science Books presented by Sapling Leaming enera emis ition Donald McQuarrie Peter A. Rock Ethan Gallogly Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide Tip: If you need to clear your work and reset the equation, click the button that looks like two red a Balance the equation AI 20,(s)-NaOH(1)-HF(g) Na,Al F6-H2Og) rrows If 12.6 kilograms of Al203(s), 57.4 kilograms of NaOH(), and 57.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced? Number kg Na, AlH Which reactants will be in excess? What is the total mass of the excess reactants left over after the reaction is complete? Al2O3 Number NaOH HF kg Previous Give Up & View Solution Check AnswerNext Exit

Explanation / Answer

a)

mol of Al2O3 --> mass/MW = 12600 / 101.96128 = 123.57 mol

mol of NaOH = mass/MW = 57400 / 39.98 = 1435.7 mol

mol of HF = mass/MW = 57.4/20 = 2870 mol

Note that...

ratio is:

Al2O3 + NaOH + HF --> Na3AlF6 + H2O

Al2O3 + NaOH + HF --> 2Na3AlF6 + H2O

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

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

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

this is now balanced

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

Al2O3 = 6NaOH; Al2O3 = 12 HF

NaOH = 2HF

mol of Al2O3 = 123.57 mol --> limiting reactnat

mol of NaOH 1435.7 mol

mol of HF = 2870 mol

ratio is

1 mol of Al2O3 = 2 mol of Na3AlF6

123.57 mol of Al2O3 --> 123.57 mol of Al2O should produce --> 2*123.57 = 247.14 mol of Na3AlF6

mass = mol*MW = 247.14*209.9412656 = 51884.8843804 g = 51.88 kg

b)

excess reactants are all but Al2O3, since it consumes all

NaOH, HF

c)

total mass excess:

Initial mass --> 12.6 + 57.4 + 57.4 = 127.4 kg initially

mass of H2O produced:

1 mol of Al2O3 --> 9 mol of H2O

123.57 mol --> 9*123.57 = 1112.13 mol of H2O

mas sof H2O = mol*MW = 1112.13*18 = 20018.34 g = 20.018 kg

Then

Total mass produced = water + cryolite = 20.018 +51.88 = 71.898 kg

then

Leftover = 127.4-71.898 = 55.502 kg of leftover material

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