Antacids are taken to relieve heartburn or indigestion caused by excess stomach acid. The active ingredients vary depending on the brand, but all contain bases. Each Alka-Seltzer(R) tablet contains 1.916 g of the active ingredient sodium bicarbonate (NaHCO3). When it dissolves, the sodium bicarbonate reacts with stomach acid (essentially 0.100 M HCl) according to the reaction below:
HCl(aq) + NaHCO3(aq) ----> NaCl(aq) + CO2(g) + H2O(l)
How many milliliters of stomach acid can one tablet of Alka-Seltzer(R) neutralize?
1.916/84.008= 0.0228
but I'm not sure how to figure out how much it will neutralize
mols NaHCO3 = 0.0228 = mols HCl
But M HCl = mols HCl/L HCl. You know M and mols, solve for L and convert to mL.
Well, let's do some calculations with a touch of humor, shall we?
First, we need to find out the number of moles of sodium bicarbonate (NaHCO3) present in one tablet. We know that the tablet contains 1.916 g of NaHCO3. Now, I could make a joke about baking a giant cake with that amount, but let's save that for another time.
Using the molar mass of NaHCO3, which is approximately 84 g/mol, we can calculate the number of moles:
1.916 g / 84 g/mol = 0.0228 mol
Now, if each mole of NaHCO3 can react with one mole of HCl, we can say that 0.0228 mol of NaHCO3 can react with 0.0228 mol of HCl. Since the concentration of HCl is 0.100 M (which stands for molarity, not "midnight snacking")...
0.0228 mol / 0.100 mol/L = 0.228 L
Now, we know that 0.228 L is equivalent to 228 mL, so one tablet of Alka-Seltzer can neutralize approximately 228 milliliters of stomach acid.
Keep in mind that this calculation assumes 100% efficiency, so results may vary depending on individual circumstances. And always remember, laughter is the best antacid.
To find out how many milliliters of stomach acid one tablet of Alka-Seltzer(R) can neutralize, we need to use stoichiometry.
First, we need to calculate the number of moles of sodium bicarbonate (NaHCO3) present in one tablet. We are given that each tablet contains 1.916 g of NaHCO3.
The molar mass of NaHCO3 is:
Na: 22.99 g/mol
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Adding these up:
22.99 + (1.01 * 1) + (12.01 + 3 * 16.00) = 84.01 g/mol
Now, we can calculate the moles of NaHCO3 in one tablet:
1.916 g / 84.01 g/mol = 0.02283 mol
According to the balanced equation, 1 mole of NaHCO3 reacts with 1 mole of HCl. Therefore, 0.02283 mol of NaHCO3 will react with the same number of moles of HCl.
Since we know the molarity (0.100 M) of HCl, we can use the formula:
moles = volume (in L) * molarity (in mol/L)
Rearranging the formula to solve for volume:
volume (in L) = moles / molarity
volume (in L) = 0.02283 mol / 0.100 mol/L = 0.2283 L
To convert liters to milliliters, we multiply by 1000:
volume (in mL) = 0.2283 L * 1000 mL/L = 228.3 mL
Therefore, one tablet of Alka-Seltzer(R) can neutralize approximately 228.3 milliliters of stomach acid.
each mole of HCl reacts with one mole of NaHCO3
so, how many moles of NaHCO3 do you have?
How many mL of 0.1M acid does that take?