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SOVA2 [1]
2 years ago
14

If 16.9 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc

ed
Chemistry
1 answer:
hodyreva [135]2 years ago
8 0

Answer: 69.72 kg of cryolite will be produced.

Explanation:

The balanced chemical equation is:

Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of Al_2O_3 = \frac{16.9\times 1000g}{102g/mol}=165.7moles

moles of NaOH = \frac{57.4\times 1000g}{40g/mol}=1435moles

moles of HF = \frac{57.4\times 1000g}{20g/mol}=2870moles

As 1 mole of Al_2O_3 reacts with 6 moles of NaOH

166 moles of  Al_2O_3 reacts with = \frac{6}{1}\times 166=996 moles of NaOH

As 1 mole of Al_2O_3 reacts with 12 moles of HF

166 moles of  Al_2O_3 reacts with = \frac{12}{1}\times 166=1992 moles of HF

Thus Al_2O_3 is the limiting reagent.

As 1 mole of Al_2O_3 produces = 2 moles of cryolite

166 moles of  Al_2O_3 reacts with = \frac{2}{1}\times 166=332 moles of cryolite

Mass of cryolite (Na_3AlF_6) = moles\times {\text {molar mass}}=332mol\times 210g/mol=69720g=69.72kg

Thus 69.72 kg of cryolite will be produced.

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