Chemistry
posted by Brit on .
Stomach acid is essentially 0.10 M HCl. An active ingredient found in a number of popular antacids is CaCO3. Calculate the number of CaCO3 needed to exactly react with 250 mL of stomach acid.
CaCO3(s)+ 2HCl(aq)= CO2(g)+ CaCl2(aq)+ H20(l)

"Calculate the number of WHAT of CaCO3 needed....."
mols HCl = M x L = ?
mols CaCO3 = 1/2 that (from the coefficients in the balanced equation).
If you want g CaCO3, then g = mols CaCO3 x molar mass CaCO3. 
multiply ch22+kks2=________________ (G)
THANK YOU! 
The answer is 1.3 grams of CaCO3. I am not getting that by multiplying mols CaCO3 x molar mass CaCO3.

Then you must be pushing the wrong buttons on the calculator.
mols HCl = M x L = 0.1 x 0.250 = 0.0125.
g CaCO3 = 0.0125 x 100.09 = 1.251 g which I would round to 1.2 g; apparently the author of the problem rounds another way to obtain 1.3 g.
I don't understand, if you tried to work this yourself, how you ended up with anything different.