Determine the concentration of a solution of Mg(NO3)2 given the precipitate formed is 2.97 g and there are 50mL of Mg(NO3)2 and 50 mL of NaOH. Please show all work for each step clearly
Mg(NO3)2 + 2NaOH ==> Mg(OH)2 + 2NaNO3
mols Mg(OH)2 formed = 2.97/molar mass Mg(OH)2.
1 mol Mg(OH)2 was formed from 1 molo Mg(NO3)2.
M Mg(NO3)2 = mols Mg(NO3)2/L soln
Note: 50 mL each soln = 100 mL total or 0.1 L.
This calculation is based upon Mg(NO3)2 and NaOH being in stoichiometric proportions; i.e. no common ion effect due to an excess of either.
To determine the concentration of the Mg(NO3)2 solution, we need to follow a few steps:
Step 1: Find the moles of the precipitate.
Given that the mass of the precipitate is 2.97 g, we can calculate the number of moles using the molar mass of Mg(NO3)2.
The molar mass of Mg(NO3)2 is:
Mg: 24.31 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (3 atoms)
Total molar mass: 24.31 + (14.01 * 2) + (16.00 * 6) = 148.33 g/mol
Number of moles = mass / molar mass
Number of moles = 2.97 g / 148.33 g/mol = 0.02 moles
Step 2: Determine the limiting reactant.
To find the limiting reactant, we compare the moles of Mg(NO3)2 with the moles of NaOH. Since there is a 1:1 mole ratio between Mg(NO3)2 and NaOH, we can use the stoichiometry to compare.
Given that both the Mg(NO3)2 solution and NaOH are 50 mL, we need to convert them to moles.
First, we need to determine the molarity of the NaOH solution. We can do this by calculating the moles of NaOH using the molarity equation:
Concentration (molarity) = moles / volume (in liters)
Since the volume is given in mL, we need to convert it to liters:
50 mL = 50/1000 = 0.05 L
Now, let's calculate the moles of NaOH:
Concentration (molarity) = moles / volume
Moles = concentration * volume
Moles = 0.1 M (given) * 0.05 L = 0.005 moles
Now, we compare the moles of Mg(NO3)2 and NaOH to determine the limiting reactant. Since there is a 1:1 mole ratio, their moles should be equal if they are the limiting reactant.
Moles of Mg(NO3)2 = 0.02 moles
Moles of NaOH = 0.005 moles
The moles of NaOH (0.005 moles) is less than the moles of Mg(NO3)2 (0.02 moles). Therefore, NaOH is the limiting reactant.
Step 3: Calculate the concentration of the Mg(NO3)2 solution.
Since NaOH is the limiting reactant, we can use the moles of NaOH to determine the moles of Mg(NO3)2, assuming a 1:1 ratio.
The moles of Mg(NO3)2 = moles of NaOH = 0.005 moles
Now, we need to convert moles to concentration using the volume of the Mg(NO3)2 solution.
Concentration = moles / volume
Volume = 50 mL = 50/1000 = 0.05 L
Concentration = 0.005 moles / 0.05 L = 0.1 M
Therefore, the concentration of the Mg(NO3)2 solution is 0.1 M.
To determine the concentration of a solution of Mg(NO3)2, we need to follow several steps. Here's how you can calculate it:
Step 1: Write down the balanced chemical equation for the reaction between Mg(NO3)2 and NaOH.
Mg(NO3)2 + 2NaOH → Mg(OH)2 + 2NaNO3
Step 2: Determine the molar mass of Mg(OH)2.
Mg = 24.31 g/mol + O = 16.00 g/mol + H = 1.01 g/mol × 2
Mg(OH)2 = 24.31 g/mol + 2(16.00 g/mol + 1.01 g/mol) = 58.33 g/mol
Step 3: Calculate the number of moles of Mg(OH)2 formed using the weight of the precipitate.
moles of Mg(OH)2 = weight of precipitate / molar mass
moles of Mg(OH)2 = 2.97 g / 58.33 g/mol = 0.051 moles
Step 4: Convert the volume of the solution to the number of moles of Mg(NO3)2.
Since we have a 1:1 mole ratio between Mg(NO3)2 and Mg(OH)2, the number of moles of Mg(NO3)2 is also 0.051 moles.
Step 5: Calculate the concentration of Mg(NO3)2.
Concentration (mol/L) = moles / volume (L)
Since we have 50 mL of Mg(NO3)2 and 50 mL of NaOH, the total volume is 100 mL or 0.1 L.
Concentration (mol/L) = 0.051 moles / 0.1 L = 0.51 mol/L
Therefore, the concentration of the Mg(NO3)2 solution is 0.51 mol/L.