Question

2.(3) Translate the following statement into a balanced chemical equation.

Three moles of calcium chloride react with two moles of sodium phosphate to produce solid calcium phosphate and six moles of sodium chloride.

3.(3) 50.0 mL of 1.0 M Ag(NO3) is added to 5.55 g of CaCl2 (MM=111.0 g). What is the concentration of nitrate ion in the final solution in moles/liter?

3a.(3) How many grams of silver chloride can be produced in this reaction?

Answer 1

(2) balanced equation

3 CaCl2 + 2 Na3PO4 --------------------------->   Ca3(PO4)2 + 6 NaCl

(3)

moles of AgNO3 = 50 x 1.0 / 1000 = 0.05

moles of CaCl2 = mass / molar mass = 5.55 / 111 = 0.05

2 AgNO3 +    CaCl2 --------------------> 2 AgCl     + Ca(NO3)2

2                       1                                      2                   1

0.05                   0.05

limiting reagent is AgNO3 so AgNO3 totally consumed

2mol AgNO3 ------------------> 1 mol Ca(NO3)2

0.05 mol -----------------------> 0.05 x 1/ 2 = 0.025 mol Ca(NO3)2

moles of NO3- = 2 x 0.025 = 0.05 mol

concentration of NO3- = moles / total volume

                                     = 0.05 / 50

                                    = 0.001M

2 AgNO3 +    CaCl2 --------------------> 2 AgCl     + Ca(NO3)2

2 mol AgNO3 gives---------------> 2mol AgCL

0.05 mol AgNO3 gives ------------> 0.05 mol AgCl

moles = mass / molar mass

0.05 = mass / 143.32

mass of AgCl = 0.05 x 143.32

mass of AgCl = 7.17g