A) A solution was prepared by dissolving 1.113 g of MgCl2 into water and made up to 50.0 ml, calculate the molar concentration of the solution.
B) A 250.0 ml of 0.50 M solution of Na3PO2 solution was prepared by diluting 2.5 M of stock solution. Calculate the volume of stock solution required.
C) Solution in (a) was mixed with 50.0 ml of diluted solution in (b), a white preciptate was formed according to the following unbalanced equation:
MgCl2 + Na3PO4---> Mg3(PO4)2 + NaCl
Calculate the mass of Mg3(PO4)2 formed.
I had done part(A) and (B), but I don't know how to do part (C). Thanks for help
Push
To solve part (C), you need to determine the limiting reactant and calculate the mass of Mg3(PO4)2 formed.
Step 1: Calculate the moles of MgCl2 and Na3PO4 in the solutions.
- In part (A), you calculated the molar concentration of the MgCl2 solution. Let's call it [MgCl2].
- In part (B), you calculated the volume of the stock solution required. Let's call it V_stock and the concentration of the stock solution C_stock.
The moles of MgCl2 in the solution can be calculated using the molar concentration ([MgCl2]) and the volume (V) in liters:
n_MgCl2 = [MgCl2] * V
The moles of Na3PO4 in the solution can be calculated using the molar concentration of the diluted solution ([Na3PO4]) and the volume (V) in liters:
n_Na3PO4 = [Na3PO4] * V
Step 2: Determine the stoichiometry of the reaction.
From the unbalanced equation:
MgCl2 + Na3PO4 ---> Mg3(PO4)2 + NaCl
We can see that 1 mole of MgCl2 reacts with 1 mole of Na3PO4 to form 1 mole of Mg3(PO4)2.
Step 3: Determine the limiting reactant.
To determine the limiting reactant, compare the ratio of moles of MgCl2 and Na3PO4. The reactant with the smaller value is the limiting reactant.
limiting reactant = min(n_MgCl2, n_Na3PO4)
Step 4: Calculate the moles of Mg3(PO4)2 formed.
Since 1 mole of MgCl2 reacts with 1 mole of Mg3(PO4)2, the moles of Mg3(PO4)2 formed is equal to the moles of the limiting reactant.
n_Mg3(PO4)2 = limiting reactant
Step 5: Calculate the mass of Mg3(PO4)2 formed.
To calculate the mass of Mg3(PO4)2 formed, you need to know the molar mass of Mg3(PO4)2. Let's call it MM_Mg3(PO4)2.
mass_Mg3(PO4)2 = n_Mg3(PO4)2 * MM_Mg3(PO4)2
By following these steps, you will be able to calculate the mass of Mg3(PO4)2 formed.