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Calculations: Solid reactants, Liquid reactants


                                      and Theoretical yield























                     ⸪ One mole of any substance = Molecular weight in grams

           Amount of A                     Amount of B                        Amount of C (grams)
              (grams)                            (ml)                           (Theoretical Yield)

           1 mole → 10 g                    1 mole → 20 g                Calculation of theoretical yield:
                                                                         1.  Balance  the  chemical  equation
           0.001 mole → ?                  0.002 mole → ?
                                                                            in terms of moles.
          ⸫ (A) Amount =                   Amount = 0.04 g

                0.01 g                      As B is liquid,              2.  Determine the limiting reactant;
                                                                            the reactant used up first.
        ◼  A is the limiting                             m (grams)

           reactant which is      ⸪ B density = ρ =           (    )     3.  Number of moles of the
             used up first.                      = 0.7 g/ml                 limiting reactant = Number of

                                                 m (g)      0.04            moles of the product.
                                  ⸫    (    ) =          =
                                                 ρ (ml)      0.7                     1 mole → 48 g

                                     ⸫                 =   .                        0.001 mole → ?

                                                                         ⸫ (C) Amount (theoretical yield) =

                                                                                         0.048 g













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