Exactly 50.0ml solution of Na2CO3 was titrated with 65.8ml of 3.00M HCl. If the specific gravity of the Na2CO3 solution is 1.25, what percent by weight of Na2CO3 does it contain?
I assume you titrated ALL of the Na2CO3; i.e., Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
mols HCl = M x L = 3.00 x 0.0658 = approximately 0.2 (you can do it more accurately).
mols Na2CO3 = 1/2 that (from the coefficients) = approximately 0.1
g Na2CO3 = mols x molar mass = 0.1 x 106 = approximately 10 g.
mass of the 50 mL soln = 50 x 1.25 = 62.5 grams.
% Na2CO3 = (10/62.5)*10 = ?%
..
To determine the percent by weight of Na2CO3 in the solution, follow these steps:
Step 1: Calculate the moles of HCl used in the titration.
The volume of 3.00M HCl used is 65.8 ml.
Moles of HCl = Volume (in L) × concentration (in mol/L)
Convert 65.8 ml to L: 65.8 ml ÷ 1000 = 0.0658 L
Moles of HCl = 0.0658 L × 3.00 mol/L = 0.1974 mol
Step 2: Convert moles of HCl to moles of Na2CO3.
HCl and Na2CO3 react in a 1:1 ratio according to the balanced chemical equation:
HCl + Na2CO3 -> NaCl + H2O + CO2
So, moles of HCl = moles of Na2CO3
Moles of Na2CO3 = 0.1974 mol
Step 3: Calculate the grams of Na2CO3 in the solution.
Molar mass of Na2CO3 = 2(Na) + 1(C) + 3(O)
= (2 × 23) + 12 + (3 × 16)
= 106 g/mol
Mass of Na2CO3 = Moles × Molar mass
= 0.1974 mol × 106 g/mol
= 20.8674 g
Step 4: Calculate the weight percent of Na2CO3 in the solution.
Weight percent of Na2CO3 = (Mass of Na2CO3 ÷ Mass of solution) × 100
Mass of solution = Volume (in ml) × specific gravity (in g/ml)
= 50.0 ml × 1.25 g/ml
= 62.5 g
Weight percent of Na2CO3 = (20.8674 g ÷ 62.5 g) × 100
= 33.39%
Therefore, the Na2CO3 solution contains approximately 33.39% Na2CO3 by weight.
To determine the percent by weight of Na2CO3 in the solution, we need to calculate the concentration of Na2CO3 in the solution and then convert it to a weight percentage.
First, let's calculate the number of moles of HCl used in the titration. We know the volume and concentration of HCl:
Volume of HCl (VHCl) = 65.8 mL
Concentration of HCl (CHCl) = 3.00 M
Using the formula:
moles of HCl = VHCl * CHCl
moles of HCl = 65.8 mL * (1 L / 1000 mL) * 3.00 mol/L
moles of HCl = 0.1974 mol
According to the balanced chemical equation for the reaction between Na2CO3 and HCl:
2 Na2CO3 + 2 HCl -> 2 NaCl + H2O + CO2
We can see that the mole ratio between Na2CO3 and HCl is 2:1. Therefore, the number of moles of Na2CO3 in the solution is twice the number of moles of HCl used.
moles of Na2CO3 = 2 * moles of HCl
moles of Na2CO3 = 2 * 0.1974 mol
moles of Na2CO3 = 0.3948 mol
To calculate the concentration of Na2CO3 in the solution, we need to divide the moles of Na2CO3 by the volume of the solution in liters:
Concentration of Na2CO3 = moles of Na2CO3 / Volume of Solution
Since we know the specific gravity of the solution, we can convert the volume to mass:
Volume of Solution = 50.0 mL = 0.05 L
Mass of Solution = Volume of Solution * Specific Gravity of the Solution
Mass of Solution = 0.05 L * 1.25 g/mL
Mass of Solution = 0.0625 g
Finally, we can calculate the weight percentage of Na2CO3 in the solution:
Weight Percentage of Na2CO3 = (Mass of Na2CO3 / Mass of Solution) * 100%
To find the mass of Na2CO3, we can use the formula:
Mass of Na2CO3 = Moles of Na2CO3 * Molar Mass of Na2CO3
The molar mass of Na2CO3 is calculated as follows:
Molar Mass of Na2CO3 = (2 * Atomic Mass of Na) + Atomic Mass of C + (3 * Atomic Mass of O)
Molar Mass of Na2CO3 = (2 * 22.99 g/mol) + 12.01 g/mol + (3 * 16.00 g/mol)
Molar mass of Na2CO3 = 105.99 g/mol
Now, we can find the mass of Na2CO3:
Mass of Na2CO3 = 0.3948 mol * 105.99 g/mol
Mass of Na2CO3 = 41.66 g
Weight Percentage of Na2CO3 = (41.66 g / 0.0625 g) * 100%
Weight Percentage of Na2CO3 = 6665.6%
Therefore, the percent by weight of Na2CO3 in the solution is approximately 6665.6%.