Cryolite, Na, AIF,(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide.Balance the equation for the synthesis of cryolite.equation: Al,0,(s) + NaOH(1) + HF(g) → Na,AlF, + H,0(g)If 15.6 kg of Al,0,(s), 57.4 kg of NaOH(1), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite willbe produced?kg Naz AIF6mass of cryolite produced:Which reactants will be in excess?HFNaOHO Al,O3What is the total mass of the excess reactants left over after the reaction is complete?total mass of excess reactants:kg

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Asked Dec 5, 2019
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Cryolite, Na, AIF,(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide.
Balance the equation for the synthesis of cryolite.
equation: Al,0,(s) + NaOH(1) + HF(g) → Na,AlF, + H,0(g)
If 15.6 kg of Al,0,(s), 57.4 kg of NaOH(1), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will
be produced?
kg Naz AIF6
mass of cryolite produced:
Which reactants will be in excess?
HF
NaOH
O Al,O3
What is the total mass of the excess reactants left over after the reaction is complete?
total mass of excess reactants:
kg
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Cryolite, Na, AIF,(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation for the synthesis of cryolite. equation: Al,0,(s) + NaOH(1) + HF(g) → Na,AlF, + H,0(g) If 15.6 kg of Al,0,(s), 57.4 kg of NaOH(1), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produced? kg Naz AIF6 mass of cryolite produced: Which reactants will be in excess? HF NaOH O Al,O3 What is the total mass of the excess reactants left over after the reaction is complete? total mass of excess reactants: kg

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Expert Answer

Step 1

The balanced chemical reaction is as follows,

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Al,O, (s) +6N2OH(1)+12HF(g)→2N ,AlF, + 9H,O(g)

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Step 2

Next, the given grams of reactant are converted into moles using mass and moles conversion.

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Mass Moles Molar mass 15.6×10° g Al,O, Moles 101.9614 g/mol =152.9 moles 57.4x10° g NaOH Moles 39.9971g / mol = 1435moles 57.4×10° g HF Moles = 20.0064 g/mol = 2869.1 moles

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Step 3

Observing the balanced chemical equation clearly shows that 1 mole of Al2O3 requires 6 moles of NaOH and 12 moles of HF. Therefore, 152.9 moles of Al2O3 requires 917.4 NaOH and 1834.8 HF moles. Hence, the limtin...

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1 mole of Al,0, =2 moles of cryolite Mass = Moles × Molar mass =(152.9x2)x 209.94 g/mol =64199.652 g =64.2 kg of cryolite

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