Answer is: mass of sodium phosphate must be added is 32.79 grams.
Balanced chemical reaction 1:
3CaCl₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₃(s) + 6NaCl(aq).
Balanced chemical reaction 2:
3Mg(NO₃)₂(aq) + 2Na₃PO₄(aq) → Mg₃(PO₄)₃(s) + 6NaNO₃(aq).
c(CaCl₂) = 0.055 M; molarity of calcium chloride.
V(CaCl₂) = 2 L; volume of calcium chloride solution.
n(CaCl₂) = c(CaCl₂) · V(CaCl₂).
n(CaCl₂) = 0.055 mol/L · 2 L.
n(CaCl₂) = 0.11 mol; amount of calcium chloride.
From balanced chemical reaction 1: n(CaCl₂) : n(Na₃PO₄) = 3 : 2.
n₁(Na₃PO₄) = 2 · 0.11 mol ÷ 3.
n₁(Na₃PO₄) = 0.073 mol; amount of sodium phosphate in reaction 1.
2) c(Mg(NO₃)₂) = 0.095 M; amount of magnesium nitrate.
V(Mg(NO₃)₂) = 2 L; volume of magnesium nitrate solution.
n(Mg(NO₃)₂) = c(Mg(NO₃)₂) · V(Mg(NO₃)₂).
n(Mg(NO₃)₂) = 0.095 mol/L · 2 L.
n(Mg(NO₃)₂) = 0.19 mol; amount of magnesium nitrate.
From balanced chemical reaction 2: n(Mg(NO₃)₂) : n(Na₃PO₄) = 3 : 2.
n₂(Na₃PO₄) = 2 · 0.19 mol ÷ 3.
n₂(Na₃PO₄) = 0.126mol; amount of sodium phosphate in reaction 2.
3) n(Na₃PO₄) = n₁(Na₃PO₄) + n₂(Na₃PO₄).
n(Na₃PO₄) = 0.073 mol + 0.126 mol.
n(Na₃PO₄) = 0.2 mol; amount of sodium phosphate.
M(Na₃PO₄) = 3Ar(Na) + Ar(P) + 4Ar(O) · g/mol.
m(Na₃PO₄) = n(Na₃PO₄) · M(Na₃PO₄).
m(Na₃PO₄) = 0.2 mol · 163.94 g/mol.
n(Na₃PO₄) = 32.788 g; mass of sodium phosphate.
Hard water has a high content of dissolved calcium and magnesium salts. 32.79 gm of sodium phosphate is needed to be added to 2.0 L of the solution.
Mass is the multiplication product of moles and the molar mass of the compound.
The balanced chemical reactions are shown as,
[tex]\rm 3CaCl_{2}(aq) + 2Na_{3}PO_{4}(aq) \rightarrow Ca_{3}(PO_{4})_{3}(s) + 6NaCl(aq)\;\; (reaction\; 1)[/tex]
And,
[tex]\rm 3Mg(NO_{3})_{2}(aq) + 2Na_{3}PO_{4}(aq) \rightarrow Mg_{3}(PO_{4})_{3}(s) + 6NaNO_{3}(aq) \;\; (reaction\; 2)[/tex]
Given,
Molarity of calcium chloride (c) = 0.055 M
Volume of calcium chloride (V) = 2.0 L
Moles of calcium chloride are calculated as:
[tex]\begin{aligned} \rm moles (n) &= \rm molarity \times volume\\\\ &= 0.055 \times 2.0 \\\\&= 0.11 \;\rm mol\end{aligned}[/tex]
From reaction 1 moles of calcium chloride: moles of trisodium phosphate = 3:2.
Moles of trisodium phosphate will be 0.073 moles.
In reaction 2 the moles of magnesium nitrate is calculated as:
Given,
Molarity of magnesium nitrate (c) = 0.095 M
Volume of magnesium nitrate (V) = 2.0 L
[tex]\begin{aligned}\rm moles (n) &= \rm molarity \times volume \\\\&= 0.095\times 2.0 \\\\&= 0.19 \;\rm mol\end{aligned}[/tex]
From reaction 2 moles of magnesium nitrate: moles of trisodium phosphate = 3:2.
Moles of trisodium phosphate will be 0.126 moles.
Total moles of sodium phosphate are: 0.073 mol + 0.126 mol = 0.2 mol.
Mass of sodium phosphate is calculated as:
[tex]\begin{aligned}\rm mass &= \rm moles \times molar\; mass\\\\&= 0.2 \times 163.94\\\\&= 32.78\;\rm g\end{aligned}[/tex]
Therefore, 32.78 g mass must be added to the solution.
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