Answer
The mass of sodium phosphate necessary is equal to 22 g.
Work Step by Step
1. Calculate the total amount of moles of Calcium chloride and Magnesium nitrate.
$$1.5 \space L \times \frac{0.050 \space mol \space CaCl_2}{1 \space L} = 0.075 \space mol \space CaCl_2 $$
$$1.5 \space L \times \frac{0.085 \space mol \space Mg(NO_3)_2}{1 \space L} = 0.12\underline{75} \space mol \space Mg(NO_3)_2 $$
2. Write the balanced reactions between these salts and sodium phosphate.
$$CaCl_2(aq) + Na_3PO_4(aq) \longrightarrow Ca_3(PO_4)_2(s) + NaCl(aq)$$ $$3CaCl_2(aq) + 2Na_3PO_4(aq) \longrightarrow Ca_3(PO_4)_2(s) + 6NaCl(aq)$$
$$Mg(NO_3)_2(aq) + Na_3PO_4(aq) \longrightarrow Mg_3(PO_4)_2(s) + NaNO_3(aq)$$ $$3 Mg(NO_3)_2(aq) + 2 Na_3PO_4(aq) \longrightarrow Mg_3(PO_4)_2(s) + 6 NaNO_3(aq)$$
- Therefore, for each 3 moles of $Mg(NO_3)_2$, we need 2 moles of sodium phosphate. And for each 3 moles of $CaCl_2$ we also need 2 moles of sodium phosphate.
3. Determine the amount of sodium phosphate necessary.
$$0.075 \space mol \space CaCl_2 \times \frac{2 \space moles \space Na_3PO_4}{3 \space moles \space CaCl_2} = 0.050 \space mol \space Na_3PO_4$$
$$0.12\underline{75} \space mol \space Mg(NO_3)_2 \times \frac{2 \space moles \space Na_3PO_4}{3 \space moles \space Mg(NO_3)_2} = 0.085 \space mol \space Na_3PO_4$$
Total = 0.050 + 0.085 = 0.135 mol $Na_3PO_4$
- Calculate or find the molar mass for $ Na_3PO_4 $:
$ Na_3PO_4 $ : ( 22.99 $\times$ 3 )+ ( 16.00 $\times$ 4 )+ ( 30.97 $\times$ 1 )= 163.94 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.135 \space mole \times \frac{ 163.94 \space g}{1 \space mole} = 22 \space g$$
