3 MgCl2(aq) + 2 Na3PO4(aq) ==> Mg3(PO4)2(s) + 6 NaCl(aq)
How many milliliters of 0.0800 M Na3PO4(aq) will react with 0.060 mol of MgCl2(aq)?
0.060 mol MgCl2 x (2 mols Na3PO4/3 mol MgCl2) = 0.060 x (2/3) = mols Na3PO4.
Then M = mols/L. You know mols and M, solve for L and convert to mL.
To determine the number of milliliters of 0.0800 M Na3PO4(aq) that will react with 0.060 mol of MgCl2(aq), we need to use the stoichiometry of the balanced chemical equation.
The balanced equation is:
3 MgCl2(aq) + 2 Na3PO4(aq) ==> Mg3(PO4)2(s) + 6 NaCl(aq)
From the balanced equation, we can see that the stoichiometric ratio between MgCl2 and Na3PO4 is 3:2, meaning that 3 moles of MgCl2 react with 2 moles of Na3PO4.
First, let's calculate the number of moles of Na3PO4 needed to completely react with 0.060 mol of MgCl2:
Number of moles of Na3PO4 = (0.060 mol MgCl2) x (2 mol Na3PO4 / 3 mol MgCl2)
= 0.040 mol Na3PO4
Next, using the molarity equation:
Molarity (M) = Moles / Volume (in liters)
We can rearrange the equation to solve for the volume (in liters):
Volume (in liters) = Moles / Molarity
Now, we substitute the values into the equation:
Volume (in liters) = 0.040 mol Na3PO4 / 0.0800 M Na3PO4
To get the volume in milliliters, we multiply by 1000:
Volume (in milliliters) = 0.040 mol Na3PO4 / 0.0800 M Na3PO4 * 1000 mL/L
Calculating the values:
Volume (in milliliters) = 500 mL
Therefore, 500 milliliters of 0.0800 M Na3PO4(aq) will react with 0.060 mol of MgCl2(aq).