chemistry
posted by Jordin on .
A 0.2528 g sample of a mixture containing sodium carbonate and sand requires 12.32 mL of 0.2409 M HCl(aq). Assuming that the sand does not react with the HCl(aq), what percentage of the mixture is sodium carbonate?
Balanced equation:
Na2CO3(s) + 2 HCl(aq) ¨ 2 NaCl(aq) + CO2(g) + H2O(l)

moles HCl = M x L = ??
moles Na2CO3 = 1/2 moles HCl (from the equation).
g Na2CO3 = moles x molar mass
%Na2CO3 = (mass Na2CO3/mass sample)*100 = ?? 
Thank you! So, is this correct?
moles HCl = M x L
= .2409 X .01232 = .00297
moles Na2CO3 = 1/2 moles HCl (from the equation).
= .00297 / 2 = .00148
g Na2CO3 = moles x molar mass
.00148 X 106 = .157
%Na2CO3 = (mass Na2CO3/mass sample)*100 = ??
.157 / .2528 X 100 = 60.8% 
Almost but not quite.
1. When multiplying 0.2409 x 0.01232, the answer is 0.00296788. There are two things wrong here. First, you threw away at least one significant figure (both numbers you multiplied have four figures so you are allowed four in the answer but you have only three). I leave all of that in the calculator and use it as is in the next calculation. You COULD round at this point to 0.002968 but I recommend you leave all of the numbers in the calculator and go to the next step.
You need to go through and correct the other steps.
2. I think you punched in the wrong numbers for the last step because
(0.157/0.2528)*100 = = 62.10%.
If we go through it all at once as follows, I get this
(0.2409 x 0.01232 x 1/2 x 105.99/0.2528)*100 = 62.21647 which rounds to 62.22% to four significant figures. 
Oh ok, I see what I did now. Thank you for the help! : )