Calculate the % Na2CO3 and % BaCO3 in a 0.2005g sample which requires 28.0mL of 0.9990 N acid for complete neutralization.
You need two equations and solve them simultaneously. The problem doesn't say which acid is used (and it doesn't need to do that) but I will assume HCl and work accordingly. Any acid can be used as long as you keep the N and M straight. The first equation is
g Na2CO3 + g BaCO3 = total mass in g.
The second equation is
mols HCl from Na2CO3 + mols HCl from BaCO3 = total mols HCl.
BaCO3 + 2HCl ==> BaCl2 + H2O + CO2
Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
Let X = mass Na2CO3
and Y = mass BaCO3
Then equation 1 is
X + Y = 0.2005
Equation 2 is
(2X/molar mass Na2CO3) + (2Y/molar mass BaCO3) = mols HCl = 0.028*0.9990
Solve the two equations for X and Y. Then
%Na2CO3 = (mass Na2CO3/0.2005)*100 = ?
%BaCO3 = (mass BaCO3/0.2005)*100 = ?
Post your work if you get stuck.
How to know their mass? Is is their molecular mass?
To calculate the percentages of Na2CO3 (sodium carbonate) and BaCO3 (barium carbonate) in the given sample, we need to follow a step-by-step approach. Here's how you can calculate it:
Step 1: Write the balanced chemical equation for the reaction between the acid and each carbonate salt:
Na2CO3 + 2HCl -> 2NaCl + H2O + CO2
BaCO3 + 2HCl -> BaCl2 + H2O + CO2
Step 2: Calculate the number of moles of the acid used:
Given that the acid concentration is 0.9990 N (normality) and the volume used is 28.0 mL, we can convert the volume to liters and find the number of moles:
Number of moles of acid = (0.9990 N) * (0.0280 L) = 0.027972 mol
Step 3: Use stoichiometry to relate the moles of acid to moles of each carbonate salt:
From the balanced chemical equations, we can see that 1 mole of Na2CO3 reacts with 2 moles of HCl, and 1 mole of BaCO3 reacts with 2 moles of HCl.
Step 4: Calculate the number of moles of Na2CO3 and BaCO3:
For Na2CO3:
Number of moles of Na2CO3 = (0.027972 mol HCl) * (1 mol Na2CO3/2 mol HCl) = 0.013986 mol Na2CO3
For BaCO3:
Number of moles of BaCO3 = (0.027972 mol HCl) * (1 mol BaCO3/2 mol HCl) = 0.013986 mol BaCO3
Step 5: Determine the molecular weights of Na2CO3 and BaCO3:
The molecular weight of Na2CO3 = 22.99 g/mol (Na) + 12.01 g/mol (C) + (3 * 16.00 g/mol) (O) = 105.99 g/mol
The molecular weight of BaCO3 = 137.33 g/mol (Ba) + 12.01 g/mol (C) + (3 * 16.00 g/mol) (O) = 197.33 g/mol
Step 6: Calculate the mass of Na2CO3 and BaCO3 in the sample:
For Na2CO3:
Mass of Na2CO3 = (0.013986 mol Na2CO3) * (105.99 g/mol Na2CO3) = 1.480 g Na2CO3
For BaCO3:
Mass of BaCO3 = (0.013986 mol BaCO3) * (197.33 g/mol BaCO3) = 2.755 g BaCO3
Step 7: Calculate the percentages of Na2CO3 and BaCO3 in the sample:
Percentage of Na2CO3 = (mass of Na2CO3 / mass of sample) * 100%
= (1.480 g / 0.2005 g) * 100% = 737.47%
Percentage of BaCO3 = (mass of BaCO3 / mass of sample) * 100%
= (2.755 g / 0.2005 g) * 100% = 1375.81%
However, it's important to note that the percentages obtained are significantly higher than 100%. This suggests that there may be an error in the calculations or that the sample may not consist solely of Na2CO3 and BaCO3. It is recommended to review the data and the calculations to ensure accuracy.