A is a solution of hydrochloric acid containing 3.650g of HCl in 1dm3 of solution. B is a
solution of impure sodium carbonate. 25cm3 of B require 22.7cm3 of A for
complete reaction. Given that 5.00g of impure B was dissolved in1dm3 solution,what is the
percentage by mass pf pure sodium carbonate in the impure sample <br /> <br /> (Na=23, C=12, O=16,
Cl=35.5, H=1.00)
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3.65 g HCl in 1 dm3 H2O is 3.65/36.5 = 0.1 M HCl
mols HCl = M x dm3 = 0.1 x 0.0227 dm3 = ?
Na2CO3 x 2HCl ==> 2NaCl + H2O + H2O
mols Na2CO3 = 1/2 mol HCl = ? and that's the mols Na2CO3 in 25 cc. Convert t mols Na2CO3 in 1 dm3 = ?mols Na2CO3 x 1000/25 = ?
Convert that to grams Na2CO3. g = mols x molar mass. That's the grams pure Na2CO3.
%Na2CO3 = (grams pure Na2CO3/mass impure sample)*100 = ?
Well, it seems like we have a chemistry problem on our hands! Let's break it down and see if we can find a funny solution.
First, let's figure out how many moles of HCl are in 1 dm³ of solution A. We know that the molar mass of HCl is 35.5 g/mol, so we can divide the mass of HCl (3.650 g) by its molar mass to find the number of moles. We'll call this Mole A.
Next, let's determine how many moles of HCl are required to react with 25 cm³ of the impure sodium carbonate solution B. Since we know that it takes 22.7 cm³ of A for complete reaction, we can set up a proportion: 22.7 cm³ of A is to Mole A as 25 cm³ of B is to Mole B. Solving for Mole B, we get the number of moles of HCl in 25 cm³ of B.
Now, let's find out how many moles of Na2CO3 (sodium carbonate) are in the impure sample. We know that the molar mass of Na2CO3 is 106 g/mol. So, we divide the mass of the impure sample (5.00 g) by the molar mass to find the number of moles. We'll call this Mole C.
Finally, let's determine the percentage by mass of pure sodium carbonate in the impure sample. We can set up a proportion: Mole B (moles of HCl in 25 cm³ of B) is to Mole C (moles of Na2CO3 in the impure sample) as 100% is to X%. Solving for X, we get the percentage by mass of pure sodium carbonate.
So, we've gone through all the steps, but unfortunately, I don't have a funny punchline for this one. Chemistry can be a serious business! But I hope this explanation helps you find the answer you're looking for.
To find the percentage by mass of pure sodium carbonate in the impure sample, we need to first determine the number of moles of HCl in solution A and the number of moles of Na2CO3 in solution B.
Step 1: Calculate the number of moles of HCl in solution A:
The molar mass of HCl is calculated by adding the atomic masses of hydrogen (H) and chlorine (Cl):
Molar mass of HCl = 1.00 g/mol (H) + 35.5 g/mol (Cl) = 36.5 g/mol
The number of moles of HCl in solution A can be calculated using the formula:
moles of HCl = mass of HCl / molar mass of HCl
moles of HCl = 3.650 g / 36.5 g/mol = 0.1 mol
Step 2: Calculate the number of moles of Na2CO3 in solution B:
The molar mass of Na2CO3 is calculated by adding the atomic masses of sodium (Na), carbon (C), and oxygen (O):
Molar mass of Na2CO3 = 2(23 g/mol) + 12 g/mol + 3(16 g/mol) = 106 g/mol
The number of moles of Na2CO3 in solution B can be calculated using the formula:
moles of Na2CO3 = mass of Na2CO3 / molar mass of Na2CO3
moles of Na2CO3 = 5.00 g / 106 g/mol = 0.047 mol
Step 3: Calculate the ratio of moles of HCl to Na2CO3:
Since the balanced chemical equation for the reaction is 2HCl + Na2CO3 → 2NaCl + CO2 + H2O, we need to convert the moles of HCl to moles of Na2CO3 using the stoichiometric coefficients. From the equation, we can see that 2 moles of HCl react with 1 mole of Na2CO3.
So, moles of HCl / moles of Na2CO3 = 2/1 = 2
Step 4: Calculate the number of moles of Na2CO3 in 25 cm3 of solution B:
The given volume of solution B is 25 cm3, so we need to convert it to dm3 by dividing it by 1000:
Volume of solution B = 25 cm3 / 1000 cm3/dm3 = 0.025 dm3
Using the ratio of moles of HCl to Na2CO3, we can find the moles of Na2CO3 in solution B:
moles of Na2CO3 = (moles of HCl / moles of Na2CO3) x volume of solution B
moles of Na2CO3 = 2 x 0.025 dm3 = 0.05 mol
Step 5: Calculate the mass of pure Na2CO3 in the impure sample:
The mass of pure Na2CO3 can be calculated using the formula:
mass of pure Na2CO3 = moles of Na2CO3 x molar mass of Na2CO3
mass of pure Na2CO3 = 0.05 mol x 106 g/mol = 5.3 g
Step 6: Calculate the percentage by mass of pure Na2CO3 in the impure sample:
The percentage by mass can be calculated using the formula:
percentage by mass = (mass of pure Na2CO3 / mass of impure sample) x 100
percentage by mass = (5.3 g / 5.00 g) x 100 = 106% (rounded to the nearest whole number)
Therefore, the percentage by mass of pure sodium carbonate in the impure sample is approximately 106%.
To find the percentage by mass of pure sodium carbonate in the impure sample, we need to calculate the molar mass of sodium carbonate, find the amount of pure sodium carbonate present in the impure sample, and then divide it by the total mass of the impure sample.
Step 1: Calculate the molar mass of sodium carbonate (Na2CO3):
The molar mass of sodium (Na) is 23 g/mol.
The molar mass of carbon (C) is 12 g/mol.
The molar mass of oxygen (O) is 16 g/mol.
The molar mass of sodium carbonate is then:
(2 * Na) + C + (3 * O) = (2 * 23) + 12 + (3 * 16) = 106 g/mol.
Step 2: Find the amount of pure sodium carbonate present in the impure sample:
From the given information, we know that 25 cm^3 of solution B requires 22.7 cm^3 of solution A for complete reaction. Since solution A contains hydrochloric acid and solution B contains impure sodium carbonate, the reaction can be represented as follows:
HCl + Na2CO3 → NaCl + CO2 + H2O
Since the reaction is a 1:1 ratio between HCl and Na2CO3, the amount of Na2CO3 in the 22.7 cm^3 of solution A is the same as the amount of Na2CO3 in the 25 cm^3 of solution B.
Step 3: Calculate the mass of pure sodium carbonate in the 25 cm^3 of solution B:
Using the concentration of solution A, we can find the mass of HCl in 22.7 cm^3 of solution A.
Concentration of solution A = 3.650 g of HCl in 1 dm^3 of solution.
First, convert 22.7 cm^3 to dm^3:
22.7 cm^3 = 22.7 * 10^(-3) dm^3 = 0.0227 dm^3
Next, calculate the mass of HCl in 0.0227 dm^3 of solution A:
Mass of HCl = Concentration * Volume
Mass of HCl = (3.650 g/dm^3) * 0.0227 dm^3 = 0.082955 g
Since the reaction is a 1:1 ratio between HCl and Na2CO3, the mass of Na2CO3 in the 25 cm^3 of solution B is the same as the mass of HCl in 22.7 cm^3 of solution A.
Step 4: Calculate the mass of pure Na2CO3 in the 5.00 g of impure B dissolved in 1 dm^3 solution:
Since the ratio of the solutions A and B is known (22.7 cm^3 of A for 25 cm^3 of B), we can set up a proportion to find the mass of Na2CO3 in 5.00 g of impure B:
(0.0227 dm^3 of A) / (25 cm^3 of B) = (0.082955 g of HCl) / (x g of Na2CO3)
Solving for x gives:
x = (0.0227 dm^3 / 25 cm^3) * (0.082955 g) = 0.07465912 g of Na2CO3
Step 5: Calculate the percentage by mass of pure Na2CO3 in the impure sample:
To find the percentage by mass of pure Na2CO3 in the impure sample, divide the mass of pure Na2CO3 (0.07465912 g) by the mass of the impure sample (5.00 g) and multiply by 100:
Percentage by mass = (0.07465912 g / 5.00 g) * 100 = 1.4931824%
Therefore, the percentage by mass of pure sodium carbonate in the impure sample is approximately 1.49%.