Question

Consider the following chemical reaction: 3 MgCl2 + 2 Na3PO4 6 NaCl + Mg3(PO4)2. Assume that 0.75 mol of MgCl2 and 0.65 mol of Na3PO4 are placed in a reaction vessel.

a) Verify that Na3PO4 is the excess reactant and MgCl2 is the limiting reactant.

b) How many moles of the excess reactant are left over after the reaction stops?

c) How many moles of NaCl will be produced in this reaction? (Remember—you must base this answer on how many moles of the limiting reactant that reacted.)

Answer 1

To determine the limiting reactant and the excess reactant, we need to compare the stoichiometry of the reaction with the amounts of each reactant given.

The balanced chemical equation is:

3 MgCl2 + 2 Na3PO4 -> 6 NaCl + Mg3(PO4)2

Given:

Moles of MgCl2 = 0.75 mol

Moles of Na3PO4 = 0.65 mol

a) To verify the limiting reactant, we need to calculate the moles of Na3PO4 and MgCl2 needed to react completely, based on the stoichiometry of the balanced equation.

From the equation, we can see that:

For every 3 moles of MgCl2, 2 moles of Na3PO4 are required.

Therefore, the moles of Na3PO4 required to react with 0.75 mol of MgCl2 would be:

(0.75 mol MgCl2) x (2 mol Na3PO4 / 3 mol MgCl2) = 0.5 mol Na3PO4

Since we have 0.65 mol of Na3PO4, which is greater than the required amount of 0.5 mol, Na3PO4 is the excess reactant.

b) To find the moles of the excess reactant left over, we subtract the moles of Na3PO4 that reacted from the initial moles:

0.65 mol Na3PO4 - 0.5 mol Na3PO4 = 0.15 mol Na3PO4 (left over)

c) To determine the moles of NaCl produced in the reaction, we need to calculate the moles of the limiting reactant (MgCl2) that reacted. From the balanced equation, we know that:

For every 3 moles of MgCl2, 6 moles of NaCl are produced.

Using the stoichiometry, we can calculate the moles of NaCl produced:

(0.75 mol MgCl2) x (6 mol NaCl / 3 mol MgCl2) = 1.5 mol NaCl

Therefore, 1.5 mol of NaCl will be produced in this reaction.