. (a) A solution was prepared by dissolving 1.4295 g of magnesium chloride (MgCl2) into
water and made up to 25.0 mL, calculate the molar concentration of the solution. (3 marks)
(b) A 250.0 mL of 0.50 M solution of sodium phosphate (Na3PO4) solution was prepared
by diluting 1.25 M of stock solution. Calculate the volume of stock solution required. (2 marks)
(c) Solution in (a) was mixed with 50.0 mL of diluted solution in (b), a white precipitate
was formed according to the following unbalanced equation:
MgCl2 (aq) + Na3PO4 (aq) � Mg3(PO4)2 (s) + NaCl (aq)
Calculate the mass of Mg3(PO4)2 formed.
(a) To calculate the molar concentration of the solution prepared in (a), we need to determine the number of moles of magnesium chloride (MgCl2) and then divide by the volume of the solution.
First, let's calculate the number of moles of MgCl2:
Number of moles of MgCl2 = mass of MgCl2 / molar mass of MgCl2
The molar mass of MgCl2 can be calculated using the atomic masses of magnesium (Mg) and chlorine (Cl):
Molar mass of MgCl2 = (molar mass of Mg x number of Mg atoms) + (molar mass of Cl x number of Cl atoms)
The atomic masses of Mg and Cl can be found on the periodic table. The molar mass of Mg is approximately 24.31 g/mol, and the molar mass of Cl is approximately 35.45 g/mol.
In this case, there is 1 magnesium atom (Mg) and 2 chlorine atoms (Cl) in MgCl2, so:
Molar mass of MgCl2 = (24.31 g/mol x 1) + (35.45 g/mol x 2)
Now, we can calculate the number of moles of MgCl2:
Number of moles of MgCl2 = 1.4295 g / Molar mass of MgCl2
Once we have the number of moles, we can calculate the molar concentration:
Molar concentration (in mol/L) = Number of moles of MgCl2 / Volume of solution (in L)
Convert the given volume from milliliters (mL) to liters (L):
Volume of solution (in L) = 25.0 mL / 1000 mL/L
Finally, substitute the values into the equation to calculate the molar concentration.
(b) To calculate the volume of stock solution required to prepare a 250.0 mL solution, we can use the dilution formula:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution (1.25 M)
V1 = volume of the stock solution needed (what we are solving for)
C2 = concentration of the final solution (0.50 M)
V2 = final volume of the solution (250.0 mL)
Rearranging the formula, we have:
V1 = (C2 x V2) / C1
Plug in the values and solve for V1 to find the volume of stock solution required.
(c) To calculate the mass of Mg3(PO4)2 formed, we need to first determine the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To find the limiting reagent, we compare the number of moles of each reactant to the stoichiometric ratio given in the unbalanced equation. The stoichiometric ratio tells us the ratio in which the reactants combine to form the product.
For MgCl2, we can calculate the number of moles using the molar concentration obtained in part (a), multiplied by the total volume of the solution after mixing (25.0 mL + 50.0 mL), converted to liters.
For Na3PO4, we can calculate the number of moles using the molar concentration (0.50 M) and the volume of the diluted solution (50.0 mL), converted to liters.
Next, we compare the moles of each reactant to the stoichiometric ratio to determine the limiting reagent. The reactant that has a smaller ratio compared to the stoichiometric ratio will be the limiting reagent.
Once we have identified the limiting reagent, we can use the stoichiometric ratio to calculate the moles of Mg3(PO4)2 formed. Finally, multiply the moles of Mg3(PO4)2 by its molar mass to find the mass of Mg3(PO4)2 formed.
(a) To calculate the molar concentration of the solution, we need to first determine the number of moles of magnesium chloride (MgCl2) present in the solution.
Given:
Mass of MgCl2 = 1.4295 g
Volume of solution = 25.0 mL = 0.0250 L
To find the number of moles of MgCl2, we use the formula:
Moles = Mass / Molar mass
The molar mass of MgCl2 can be calculated as follows:
Molar mass of MgCl2 = Molar mass of Mg + 2 * Molar mass of Cl
= (24.31 g/mol) + 2 * (35.45 g/mol)
= 95.21 g/mol
Now, we can calculate the number of moles of MgCl2:
Moles of MgCl2 = 1.4295 g / 95.21 g/mol
= 0.01504 mol
Finally, we can calculate the molar concentration of the solution:
Molar concentration = Moles / Volume of solution
= 0.01504 mol / 0.0250 L
= 0.6016 M
Therefore, the molar concentration of the solution is 0.6016 M.
(b) To calculate the volume of stock solution required, we can use the dilution formula:
C1V1 = C2V2
Where:
C1 = Initial concentration = 1.25 M
V1 = Initial volume = unknown
C2 = Final concentration = 0.50 M
V2 = Final volume = 250.0 mL = 0.250 L
Rearranging the formula, we get:
V1 = (C2 * V2) / C1
= (0.50 M * 0.250 L) / 1.25 M
= 0.10 L
Therefore, the volume of the stock solution required is 0.10 L.
(c) The balanced equation for the reaction is:
3 MgCl2 (aq) + 2 Na3PO4 (aq) -> Mg3(PO4)2 (s) + 6 NaCl (aq)
From the balanced equation, we can see that the ratio of moles of MgCl2 to moles of Mg3(PO4)2 is 3:1.
We already calculated the moles of MgCl2 in part (a) as 0.01504 mol.
Therefore, moles of Mg3(PO4)2 = 1/3 * 0.01504 mol
= 0.00501 mol
To calculate the mass of Mg3(PO4)2 formed, we need to use its molar mass.
The molar mass of Mg3(PO4)2 can be calculated as follows:
Molar mass of Mg3(PO4)2 = (3 * Molar mass of Mg) + (2 * Molar mass of P) + (8 * Molar mass of O)
= (3 * 24.31 g/mol) + (2 * 30.97 g/mol) + (8 * 16.00 g/mol)
= 262.86 g/mol
Now, we can calculate the mass of Mg3(PO4)2 formed:
Mass = Moles * Molar mass
= 0.00501 mol * 262.86 g/mol
= 1.32 g
Therefore, the mass of Mg3(PO4)2 formed is 1.32 g.