Cryolite, Na3AlF6 (s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation for the synthesis of cryolite. equation: Al2O3 (s) + NaOH (l) + HF (g) ⟶Na3AlF6+H2O (g) If 14.0 kg of Al2O3 (s), 59.4 kg of NaOH (l), and 59.4 kg of HF (g) react completely, how many kilograms of cryolite will be produced? mass of cryolite produced: kg Na3AlF6 Which reactants will be in excess? HF Al2O3 NaOH What is the total mass of the excess reactants left over after the reaction is complete? total mass of excess reactants: kg question source: MRG - General Chemistry|publisher: University Science Books

m (Na3AlF6) = 57.6537 kg. Leftover mass will be 52.8644 kg

Explanation:

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

n (Al2O3) = m (Al2O3) / M (Al2O3) = 14000/101.96=137.31 mol

n (NaOH) = m (NaOH) / M (NaOH) = 59400/40=1485 mol

n (HF) = m (HF) / M (HF) = 59400/20.01=2968 mol

To know what is in excess we divide moles by coefficients and compare numbers.

n (Al2O3) = 137.31 mol

n (NaOH) / 6=247.5 mol

n (HF) / 12=247.33 mol

so NaOH and HF are in excess

So

m (Na3AlF6) = n (Na3AlF6) * M (Na3AlF6) / 1000=n (Al2O3) * 2*M (Na3AlF6) / 1000=57.6537 kg.

Leftover mass will be (n (NaOH) - n (Al2O3) * 6) * 40 + (n (HF) - n (Al2O3) * 12) * 20.01=26445.6+26418.8=52864.4 g = 52.8644 kg