Asked by Summer

Calculate the % Na2CO3 and % BaCO3 in a 0.2005g sample which requires 28.0mL of 0.9990 N acid for complete neutralization.
Answered by DrBob222
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.
Answered by Summer
How to know their mass? Is is their molecular mass?
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