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.

To determine the number of CaCO3 required to react with 250 mL of stomach acid, we need to use the balanced equation and stoichiometry.

First, let's convert the volume of stomach acid from milliliters to liters:
250 mL = 250/1000 L = 0.25 L

Next, we need to find the moles of HCl in the stomach acid. We can use the molarity (0.10 M) and the volume (0.25 L) to calculate this:
moles of HCl = molarity × volume
moles of HCl = 0.10 M × 0.25 L = 0.025 moles

Looking at the balanced equation, we can see that the stoichiometric ratio between HCl and CaCO3 is 2:1. This means that 2 moles of HCl react with 1 mole of CaCO3.

So, to determine the moles of CaCO3 needed, we can use the stoichiometric ratio:
moles of CaCO3 = moles of HCl / 2
moles of CaCO3 = 0.025 moles / 2 = 0.0125 moles

Finally, to find the number of CaCO3 particles, we can use Avogadro's number.
1 mole of any substance contains 6.022 × 10^23 particles.

Therefore, the number of CaCO3 particles needed can be calculated as follows:
number of CaCO3 particles = moles of CaCO3 × Avogadro's number
number of CaCO3 particles = 0.0125 moles × 6.022 × 10^23 particles/mol

To get the final answer, simply perform the calculation:
number of CaCO3 particles = 7.52 × 10^21 particles

Therefore, approximately 7.52 × 10^21 particles of CaCO3 are needed to exactly react with 250 mL of stomach acid.

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.