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Sagot :
Answer: 31.8 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Al_2O_3=\frac{60.0g}{102g/mol}=0.59moles[/tex]
[tex]\text{Moles of} C=\frac{30.0g}{12g/mol}=2.5moles[/tex]
[tex]Al_2O_3+3C\rightarrow 2Al+3CO[/tex]
According to stoichiometry :
1 mole of [tex]Al_2O_3[/tex] require 3 moles of [tex]C[/tex]
Thus 0.59 moles of [tex]Al_2O_3[/tex] will require=[tex]\frac{3}{1}\times 0.59=1.77moles[/tex] of [tex]C[/tex]
Thus [tex]Al_2O_3[/tex] is the limiting reagent as it limits the formation of product and [tex]C[/tex] is the excess reagent as it is present in more amount than required.
As 1 mole of [tex]Al_2O_3[/tex] give = 2 moles of [tex]Al[/tex]
Thus 0.59 moles of [tex]Al_2O_3[/tex] give =[tex]\frac{2}{1}\times 0.59=1.18moles[/tex] of [tex]Al[/tex]
Mass of [tex]Al=moles\times {\text {Molar mass}}=1.18moles\times 27g/mol=31.8g[/tex]
Thus 31.8 g of [tex]Al[/tex] will be produced from the given masses of both reactants.
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