Question

If 11.8 kilograms of Al2O3(s), 57.4 kilograms of NaOH(l), and 57.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?

Answer 1

Question #47828, Chemistry, Other

$$6NaOH + Al_2O_3 + 12HF \rightarrow 2Na_3[AlF_6] + 9H_2O$$ $$\nu_{NaOH} = \frac{m_{NaOH}}{M_{NaOH}} = \frac{57.4}{40} = 1.435\ \text{kmol}$$ $$\nu_{Al_2O_3} = \frac{m_{Al_2O_3}}{M_{Al_2O_3}} = \frac{11.8}{102} = 0.115\ \text{kmol}$$ $$\nu_{HF} = \frac{m_{HF}}{M_{HF}} = \frac{57.4}{20} = 2.87\ \text{kmol}$$ $$\nu_{Na_3[AlF_6]} = 2\nu_{Al_2O_3} = 0.230\ \text{kmol}$$ $$m_{Na_3[AlF_6]} = M_{Na_3[AlF_6]} * \nu_{Na_3[AlF_6]} = 0.23\ \text{kmol} * 214\ \frac{\text{kg}}{\text{kmol}} = 49.22\ \text{kg}$$

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