Question

IF 23 g of FeCl2 reacts with 41g Na3PO4 what is the limiting reactant?

Answer 1

First, we need to figure the reaction out.

Both reactants are ionic compounds, so their reaction is probably a double displacemente reaction.

The Cl has a charge of 1-, so Fe has a charge of 2+. Na has a charge of +1 and PO₄ has a charge of 3-.

To combine Fe²⁺ with PO₄³⁻, we will need 3 Fe²⁺ and 2 PO₄³⁻, so the compound becomes neutral.

Both Na and Cl have single charge, so their combination is simply 1 Na and 1 Cl.

So, the products are Fe₃(PO₄)₂ and NaCl, so the unbalanced reaction is:

$FeCl_2+Na_3PO_4\to Fe_3(PO_4)_2+NaCl$

To balance it, We can start by putting a coefficient of 2 on Na₃PO₄, so we have two PO₄³⁻ anions on both sides:

$FeCl_2+2Na_3PO_4\to Fe_3(PO_4)_2+NaCl$

Now, we put a coefficient of 3 on FeCl₂ and 6 on NaCl to get the balanced equation:

$3FeCl_2+2Na_3PO_4\to Fe_3(PO_4)_2+6NaCl$

Now, we first need to calculate the number of moles of each reactant, using their molar masses:

$\begin{gathered} M_{FeCl_2}=1\cdot M_{Fe}+2\cdot M_{Cl}=(1\cdot55.845+2\cdot35.453)g/mol=126.751g/mol \\ M_{Na_3PO_4}=3\cdot M_{Na}+1\cdot M_P+4\cdot M_O=(3\cdot22.98976928+1\cdot30.973762+4\cdot15.9994)g/mol=163.94066984g/mol \end{gathered}$

So:

$\begin{gathered} M_{FeCl_{2}}=\frac{m_{FeCl_2}}{n_{FeCl_{2}}} \\ n_{FeCl_2}=\frac{m_{FeCl_2}}{M_{FeCl_{2}}}=\frac{23g}{126.751g/mol}=0.181458\ldots mol \end{gathered}$$\begin{gathered} M_{Na_3PO_4}=\frac{m_{Na_3PO_4}}{n_{Na_{3}PO_{4}}} \\ n_{Na_3PO_4}=\frac{m_{Na_3PO_4}}{M_{Na_{3}PO_{4}}}=\frac{41g}{163.9406698g/mol}=0.250090\ldots mol \end{gathered}$

Now, if the coefficient of FeCl₂ is 3, for each 3 FeCl₂ that reacts, one reaction occurs, so the number of reactions that will need to occur to consume all its number of moles is that amount over 3:

$n_{\text{reaction}}=\frac{n_{FeCl_2}}{3}=\frac{0.181458\ldots mol}{3}=0.060486\ldots mol$

Similarly, we get the number of reactions that would occur if all Na₃PO₄ reacts by dividing its number of moles by its coefficient, that is 2:

$n_{\text{reaction}}=\frac{n_{Na_3PO_4}}{2}=\frac{0.250090\ldots mol}{2}=0.125045\ldots mol$

This means that the amount we have of FeCl₂ is enough for only about 0.0605 mol of reactions while the amount we have of Na₃PO₄ is enought for about 0.1250 mol of reactions, which is more than the other reactant.

This means that the reactant that limits the amount of reaction that can occur is FeCl₂, so the limiting reactant is FeCl₂.