Question

For the following reaction 5.20 mol (NH4)3PO4 reacts with 4.35 moles Pb(NO3)4 according to the reaction below.Fill out the BCA table to determine the quantities of each substance throughout the reaction and use it to answer the questions 4(NH4)3PO4+3Pb(NO3)4=Pb3(PO4)4+12NH4NO3

Answer 1

Step-by-step explanation:

First, let's write the chemical reaction:

`4(NH_4)_3PO_4+3Pb(NO_3)_4\rightarrow Pb_3(PO_4)_4+12NH_4NO_3.`

Now, let's find what is the limiting reactant. Let's find how many moles of Pb3(PO4)4 are being produced by each reactant.

You can see that 4 moles of (NH4)3PO4 reacted produces 1 mol Pb3(PO4)4, so 5.20 moles of (NH4)3PO4 will produce:

`5.20\text{ moles \lparen NH}_4)_3PO_4\cdot\frac{1\text{ mol Pb}_3(PO_4)_4}{4\text{ moles \lparen NH}_4)_3PO_4}=1.3\text{ moles Pb}_3(PO_4)_4.`

And now, let's see how many moles of Pb3(PO4)4 are being produced by 4.35 moles of Pb(NO3)4. In the chemical reaction you can see that 3 moles of Pb(NO3)4 reacted produces 1 mol of Pb3(PO4)4:

`4.35\text{ moles Pb\lparen NO}_3)_4\cdot\frac{1\text{ mol Pb}_3(PO_4)_4}{3\text{ moles Pb\lparen NO}_3)_4}=1.45\text{ moles Pb}_3(PO_4)_4.`

You can realize that we have an excess of Pb(NO3)4 because the limit is imposed by (NH4)3PO4. So let's find how many moles of Pb(NO3)4 are required to produce 1.3 moles of Pb(NO3)4, like this:

`1.3\text{ moles Pb}_3(PO_4)_4\cdot\frac{3\text{ moles Pb\lparen NO}_3)_4}{1\text{ mol Pb}_3\text{\lparen PO}_4)_4}=3.9\text{ moles Pb\lparen NO}_3)_4.`

This means that we have an excess of 0.45 moles of Pb(NO3)4 (4.35 moles - 3.9 moles = 0.45 moles).

Now, let's find the number of moles that are being produced of NH4NO3 by the limiting reactant which is (NH4)3PO4. 4 moles of (NH4)3PO4 reacted produces 12 moles of NH4NO3:

`5.20\text{ moles \lparen NH}_4)_3PO_4\cdot\frac{12\text{ moles NH}_4NO_3}{4\text{ mol \lparen NH}_4)_3PO_4}=15.6\text{ moles NH}_4NO_3.`

If we follow the same steps that we followed before, we're going to see that the excess (Pb(NO3)4) will produce 17.4 moles of NH4NO3, and actually, to form 15.6 moles of NH4NO3 we need 3.9 moles of Pb(NO3)4 as we calculated before.

Answer:

And finally, we can complete the BCA table: