If 203 ML of water is added to 5.00 ML of 4.16 MKCI solution what is the concentration of the diluted solution

The answer to your question is below

Explanation:

Data

mass of CaCO₃ = 155 g

mass of HCl = 250 g

mass of CaCl₂ = 142 g

reactants = CaCO₃ + HCl

products = CaCl₂ + CO₂ + H₂O

1.- Balanced chemical reaction

             CaCO₃ + 2HCl   ⇒    CaCl₂ + CO₂ + H₂O

2.- Limiting reactant

molar mass of CaCO₃ = 40 + 12 + 48 = 100 g

molar mass of HCl = 2[1 + 35.5 ] = 73 g

theoretical proportion CaCO₃ /HCl = 100 / 73 = 1.37

experimental proportion CaCO₃ /HCl = 155 / 250 = 0.62

As the experimental proportion was lower than the theoretical proportion the limiting reactant is CaCO₃

3.-

Calculate the molar mass of CaCl₂

CaCl₂ = 40 + 71 = 111 g

          100 g of CaCO₃ 111 g of CaCl₂

           155 g of CaCO₃ x

               x = (155 x 111) / 100

               x = 17205 / 100

              x = 172.05 g of CaCl₂

4.- percent yield

Percent yield = 142 / 172.05 x 100 = 82.5 %

5.- Excess reactant

    100 g of CaCO₃   73 g of HCl

     155 g of caCO₃ x

           x = (155 x 73)/100

           x = 133.15 g

Mass of HCl = 250 - 133.15

                    = 136.9 g