posted by Fai .
A 0.682g sample of impure Na2hco3 yield a solid residue consisting of nahco3 and other solid a mass of 0.467g. What was the mass percent of nac03in sample.
I do not understand I want you step by step. Clear explain for me.
nahco3 = 84
n22hco3 = 106
You still have too many typos.
If the problem is this.
A 0.682 g sample of impure NaHCO3 yield a solid residue consisting of Na2CO3 and other solid with a mass of 0.467 g. What was the mass percent of NaHCO3?
Let x = mass NaHCO3.
and y = mass of the impurity.
That's two unknowns; therefore, you must have two equations and solve them simultaneously.
The first equation is
x + y = 0.682
The second equation is
mass Na2CO3 + mass impurity = 0.467g.
mass Na2CO3 = (x*molar mass Na2CO3/2*molar mass NaHCO3). Therefore, the two equations that must be solved are
x + y = 0.672
(106x/2*84) + y = 0.467
After you find x (which is mass NaHCO3), then
%NaHCO3 = (mass NaHCO3/mass sample(*100 = ?