Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation. If 14.5 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? Which reactants will be in excess? What is the total mass of the excess reactants left over after the reaction is complete?
AL2O3 + 6 NaOH + 12 HF --> 2 Na3AlF6 + 9 H2O
FROM EQUATION
1 mol AL2O3 = 6 mol NaOH = 12 mol HF = 2 mol Na3AlF6 = 9 mol H2O
No of mol of Al2o3 = 14.5*10^3/ 101.96 = 142.21 mol
No of mol of HF = 53.4*10^3/20 = 2670 mol
No of mol of NaOH = 53.4*10^3/40 = 1335 mol
limiting reactant is = Al2o3
No of mol of Na3AlF6 produced = 142.21*2 = 284.42 mol
mass of Na3AlF6 produced = 284.42*209.94 = 59711.14 GRAMS
= 59.711 kg
excess reactants
NaOH,HF
NaOH EXCESS = (1335-(6*142.21))*40 = 19.27 Kg
HF excess = (2670-(12*142.21))*20 = 19.27 kg
Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide....
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) -----> Na3(g)AIF6 + H2O(g) If 12.2 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? Which reactants will be in excess? What is the total mass of the excess reactants left over after the reaction is complete?
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