Stomach acid is composed primarily of HCl. If the concentration of the acid is 0.010M and it has a volume of 70 mL, how many generic tablets would you need to fully neutralize the acid? (Use your data)
I just don't know where to start for this question, or the steps to answer it.
depends on tablet size, and the contents. Dosages are available from 500mg to 1175 mg.
So we will start with the "extra Strength" 750 mg, CaCO3
moles acid: .01*.070=.007 moles
Now it takes half that number of moles of calcium carbonate ..
.0035=.750*N*/molmassCaCO3
N= .0035*mllemassCaCO3/.750
N= .0035*100/.750 = .46 tabs
Wait, could you please explain a little more in depth please? I got lost after you started with the Extra Strength.
The generic brand I'm using has 500 mg of active ingredient (CaCO3).
Also, can you explain why you took half the number of moles of CaCO3??
To answer this question, we need to consider the reaction between the stomach acid (HCl) and the generic tablets that neutralize it.
The balanced chemical equation for the reaction between HCl and the generic tablets may not be provided, so let's assume that each generic tablet contains a substance that can neutralize one mole of HCl.
We have the concentration of the stomach acid (HCl), which is 0.010M, and the volume of the acid, which is 70 mL (or 0.070 L).
To find the moles of HCl, we can use the formula:
moles = concentration x volume
moles of HCl = 0.010M x 0.070 L
Now, we have the number of moles of HCl in the given volume.
Since each generic tablet neutralizes one mole of HCl, the number of tablets needed to fully neutralize the acid is equal to the number of moles of HCl.
In other words, the number of generic tablets needed = moles of HCl.
So, to fully neutralize the acid, you would need the same number of generic tablets as the number of moles of HCl calculated above.
Let me know if you need help with the calculations.