Question

consider the reaction: 2 na₃po₄(aq) + 3 cu(no₃)₂(aq) → cu₃(po₄)₂(s) + 6 nano₃(aq). what mass of cu₃(po₄)₂ can be formed when 86.5 ml of a 0.250 m solution of na₃po₄ is mixed with 13.5 ml of a 0.420 m solution of cu(no₃)₂?

Explanation:

Step1: Calculate moles of reactants

Use the formula $n = M\times V$, where $n$ is moles, $M$ is molarity and $V$ is volume in liters.
For $Na_3PO_4$: $V_{Na_3PO_4}=86.5\ mL = 0.0865\ L$, $M_{Na_3PO_4}=0.250\ M$, so $n_{Na_3PO_4}=0.250\ mol/L\times0.0865\ L = 0.021625\ mol$.
For $Cu(NO_3)_2$: $V_{Cu(NO_3)_2}=13.5\ mL = 0.0135\ L$, $M_{Cu(NO_3)_2}=0.420\ M$, so $n_{Cu(NO_3)_2}=0.420\ mol/L\times0.0135\ L = 0.00567\ mol$.

Step2: Determine the limiting reactant

From the balanced equation $2Na_3PO_4(aq)+3Cu(NO_3)_2(aq)
ightarrow Cu_3(PO_4)_2(s)+6NaNO_3(aq)$, the mole - ratio of $Na_3PO_4$ to $Cu(NO_3)_2$ is $\frac{2}{3}$.
The moles of $Na_3PO_4$ required to react with $0.00567\ mol$ of $Cu(NO_3)_2$ is $n_{Na_3PO_4\ required}=\frac{2}{3}\times0.00567\ mol = 0.00378\ mol$. Since $0.00378\ mol<0.021625\ mol$, $Cu(NO_3)_2$ is the limiting reactant.

Step3: Calculate moles of $Cu_3(PO_4)_2$ formed

The mole - ratio of $Cu(NO_3)_2$ to $Cu_3(PO_4)_2$ is $\frac{3}{1}$. So the moles of $Cu_3(PO_4)_2$ formed, $n_{Cu_3(PO_4)_2}=\frac{1}{3}\times n_{Cu(NO_3)_2}=\frac{1}{3}\times0.00567\ mol = 0.00189\ mol$.

Step4: Calculate mass of $Cu_3(PO_4)_2$ formed

The molar mass of $Cu_3(PO_4)_2$ is $M = 3\times63.55\ g/mol+2\times(30.97\ g/mol + 4\times16.00\ g/mol)=380.59\ g/mol$.
The mass of $Cu_3(PO_4)_2$ formed, $m = n\times M=0.00189\ mol\times380.59\ g/mol\approx0.719\ g$.

Answer:

$0.719$