Question

For each of the following unbalance chemical equation suppose that exactly 15.0g of each reactant is taken. Determine which reactant is limiting and calculate what mass of each product is expected. ( assume that the limiting reactant is completely consume )

A. Al + HCl = AlCl3 + H2

B. NaOH + CO2 = Na2 CO3 + H2O

C. Pb(NO3)2 + HCl= PbCl2 + HNO 3

D. K +I 2 = KI

Answer 1

A. The balanced equation is   2Al + 6HCl 2AlCl₃ + 3H₂

From the equation,

2 moles of Al reacts with 6 moles of HCl

                       OR

2x27 g of Al reacts with 6x36.5 g of HCl

M g of Al reacts with 15.0 g of HCl

M = ( 15.0 x 2x27/ (6x36.5)

   = 3.69 g of Al

So 15.0 -3.69 = 11.31 g of Al left unreacted which is the excess reactant.

Since all the mass of HCl completly reacted HCl is the limiting reactant.

Molar mass of AlCl3 = 27 + (3x35.5) = 133.5 g/mol

Molar mass of H2 is = 2x1 = 2 g/mol

From the balanced equation ,

6 moles of HCl produces 2 mole of AlCl₃ & 3 moles of H₂

                                        OR

6x36.5 g of HCl produces 2x133.5 g of AlCl₃ & 3x2 g of H₂

15.0 g of HCl produces M g of AlCl₃ & N g of H₂

M = (15.0 x 2x 133.5 ) / (6x36.5)

   = 18.3 g of AlCl₃

N = (15.0 x 3x2 ) / (6x36.5)
   = 0.41 g of H₂

Therefore 18.3 g of AlCl₃ & 0.41 g of H₂ are formed.

Similarly do the same procedure to remaining

B. NaOH + CO2 Na2 CO3 + H2O

C. Pb(NO3)2 + HCl PbCl2 + HNO 3

D. K +I 2 KI