Question

In: Chemistry

1. When 100.0 mL of a 0.135 M Ca(NO3)2 (MM = 164.088 g/mol) solution is added...

1. When 100.0 mL of a 0.135 M Ca(NO3)2 (MM = 164.088 g/mol) solution is added to 50.0 mL of a 0.575 M K3PO4 (MM = 212.27 g/mol) solution, what mass of calcium phosphate (MM = 310.17 g/mol) will precipitate?

2. What is the final Na+ concentration if 17.0 mL of a 3.12 M NaCl solution is added to 83.0 mL of a 1.18 M Na2SO4 solution?

Solutions

Expert Solution

moles of Ca(NO3)2 = 100 x 0.135 / 1000 = 0.0135

moles of K3PO4   = 50 x 0.575 / 1000 = 0.0288

3Ca(NO3) 2 + 2K3PO4 ----------------------> Ca3(PO4)2 + 6KNO3

3                       2                                            1

0.0135             0.0288

here limiting reagent is Ca(NO3) 2 so product based on limiting reagent

3 mol Ca(NO3) 2 ------------------> 1 mol Ca3(PO4)2

0.0135 mol Ca(NO3) 2 ------------------> x mol Ca3(PO4)2

x = 0.0135 / 3 = 4.5 x 10^-3 mol precipitate

moles = mass / molar mass

4.5 x 10^-3 = mass / 310 /17

mass = 1.40 g

precipitate mass = 1.40 g

2)

[Na+] = (17 x 3.12 + 2 x 1.18 x 83 ) / (17 + 83)

            = 2.49 M

final concentration of Na+ = 2.49 M

queen_honey_blossom answered 1 hour ago
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