"2Al_2O_3(s) + 3C(s) \\rightarrow 4Al(s) +3CO_2(g)"
Moles of aluminum:
"Yield \\% =\\frac{n_{actual}}{n_{thearetical}}\\times \\%"
"\\therefore n_{theoretical} = \\frac{n_{actual}\\times 100\\%}{yield \\%}= \\frac{333.58 mol\\times 100\\%}{67.0\\%}= 497.88 mol"
According to second equation mole ratio "n(Al_2O_3):n(Al)=2:4" then "n(Al_2O_3)= \\frac{n(Al)}{2} = \\frac{497.88 mol}{2} = 248.94 mol"
"n_{theoretical}(Al_2O_3)= \\frac{n_{actual}\\times 100\\%}{yield \\%} = \\frac{248.94 mol\\times 100\\%}{77.0\\%} = 323.3 mol"
According to the first equation mole ratio "n(Al(OH)_3 : n(Al_2O_3)= 2:1" ,then
"m(Al(OH)_3)= n\\times M = 646.6 m\\times 78.00\\frac{g}{mol} = 50435 g = 50.435 kg = 50 kg"
Comments
Leave a comment