Asked by ZEE
                Find the mass of AlCl3 that is produced when 10.0 grams of Al2O3 react with 25.0 g of HCl according to the following equation.
Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(aq) 65.4 g
help me ASAP.PLIS..
            
        Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(aq) 65.4 g
help me ASAP.PLIS..
Answers
                    Answered by
            DrBob222
            
    This is a limiting reagent problem (LR) and you know that because amounts are given for both reactants.
Convert g Al2O3 to mols. mols = g/molar mass.
Convert g HCl to mols. same procedure.
Using the coefficients in the balanced equation, convert mols Al2O3 to mols AlCl3.
Do the same to convert mols HCl to mols AlCl3.
It is likely these to values will not agree; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
Now convert mols AlCl3 to g. grams = mols x molar mass AlCl3.
I don't know what the number 65.4 g means.
    
Convert g Al2O3 to mols. mols = g/molar mass.
Convert g HCl to mols. same procedure.
Using the coefficients in the balanced equation, convert mols Al2O3 to mols AlCl3.
Do the same to convert mols HCl to mols AlCl3.
It is likely these to values will not agree; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
Now convert mols AlCl3 to g. grams = mols x molar mass AlCl3.
I don't know what the number 65.4 g means.
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