Answer to Question #122129 in General Chemistry for alicia

Question

If 50.0 mL of 0.125 mol/L sodium phosphate solution is reacted with 50.0 mL of 0.125 mol/L magnesium nitrate solution, what is the predicted mass of precipitate in g?

Answer 1

  Solution.

 _{0.00625               0,00625                           y}

2Na₃PO₄ + 3Mg(NO₃)₂ → Mg₃(PO₄)₂↓ + 6NaNO₃

      ^{2                               3                               1}

V(Na₃PO₄) = 50,0 mL = 0,05 L;  

V(Mg(NO₃)₂) = 50,0 mL = 0,05 L;

n(Na₃PO₄) = c*V = 0,125*0,05 = 0,00625 mol;

n(Mg(NO₃)₂) = c*V = 0,125*0,05 = 0,00625 mol;

Let n(Mg(NO₃)₂) is equal to x, therefore x = 0,00625*3/2 = 0,009375 mol.

We have only 0,00625 mol Mg(NO3)2;

Mg(NO3)2 – in lack.

n(Mg₃(PO₄)₂) = y = 0,00625*1/3 = 0,002083 mol;

m(Mg3(PO4)2) = 0,002083*263 = 0,548 g.

 Answer: 0,548 g Mg3(PO4)2.