Stomach acid is 0.02M HCl. A 330mg antacid tablet is 41% magnesium hydroxide and 36.2% sodium Bicarbonate. What is the volume of acid that can be neutralized by one tablet?
moles of Mg(OH)2: .41*330/molmassMg(OH)2
moles of Na2(CO3): .59*330/molmassNa2CO3
moles of monotonic acid neutralized: twice the sum of above
volumeacid: molesacidabove/.02
1.512
To find the volume of acid neutralized by one tablet, we need to calculate the number of moles of acid-neutralizing compounds in the tablet.
First, find the mass of magnesium hydroxide in one tablet.
330 mg x 0.41 = 135.3 mg magnesium hydroxide
Convert the mass of magnesium hydroxide to moles.
Molar mass of magnesium hydroxide (Mg(OH)2) = 24.31 g/mol (Mg) + 2(16.00 g/mol (O) + 1.01 g/mol (H)) = 58.33 g/mol
135.3 mg = 0.1353 g
0.1353 g / 58.33 g/mol ≈ 0.00232 mol magnesium hydroxide
Next, find the mass of sodium bicarbonate in one tablet.
330 mg x 0.362 = 119.46 mg sodium bicarbonate
Convert the mass of sodium bicarbonate to moles.
Molar mass of sodium bicarbonate (NaHCO3) = 22.99 g/mol (Na) + 1.01 g/mol (H) + 12.01 g/mol (C) + 3(16.00 g/mol (O)) = 84.01 g/mol
119.46 mg = 0.11946 g
0.11946 g / 84.01 g/mol ≈ 0.00142 mol sodium bicarbonate
Now, calculate the number of moles of HCl that can be neutralized by each compound.
For magnesium hydroxide (Mg(OH)2):
1 mole of Mg(OH)2 reacts with 2 moles of HCl.
0.00232 mol magnesium hydroxide x 2 mol HCl / 1 mol magnesium hydroxide = 0.00464 mol HCl
For sodium bicarbonate (NaHCO3):
1 mole of NaHCO3 reacts with 1 mole of HCl.
0.00142 mol sodium bicarbonate x 1 mol HCl / 1 mol sodium bicarbonate = 0.00142 mol HCl
To find the total moles of HCl neutralized by one tablet, sum the moles from both compounds.
0.00464 mol HCl + 0.00142 mol HCl = 0.00606 mol HCl
Finally, calculate the volume of acid that can be neutralized by one tablet using the molarity of stomach acid.
Molarity (M) = Moles (mol) / Volume (L)
0.02 M = 0.00606 mol HCl / Volume (L)
Rearrange the equation to solve for the volume:
Volume (L) = Moles (mol) / Molarity (M)
Volume (L) = 0.00606 mol HCl / 0.02 M
Volume (L) = 0.303 L
Therefore, one tablet can neutralize approximately 0.303 liters or 303 milliliters of stomach acid.
To find the volume of acid that can be neutralized by one tablet, we need to determine how many moles of the acid are present in 330 mg of the antacid tablet.
Let's first find the mass of magnesium hydroxide and sodium bicarbonate in the tablet:
Mass of magnesium hydroxide = 41% of 330 mg = 0.41 * 330 mg = 135.3 mg
Mass of sodium bicarbonate = 36.2% of 330 mg = 0.362 * 330 mg = 119.46 mg
Next, we need to convert the mass of each substance to moles. To do this, we will use their respective molecular weights:
Molecular weight of magnesium hydroxide (Mg(OH)2) = 24.31 g/mol + 2 * 16.00 g/mol + 2 * 1.01 g/mol = 58.33 g/mol
Molecular weight of sodium bicarbonate (NaHCO3) = 22.99 g/mol + 1.01 g/mol + 3 * 16.00 g/mol = 84.01 g/mol
Moles of magnesium hydroxide = mass / molecular weight = 135.3 mg / 58.33 g/mol = 0.00232 mol
Moles of sodium bicarbonate = mass / molecular weight = 119.46 mg / 84.01 g/mol = 0.00142 mol
Now, let's determine the moles of hydrochloric acid (HCl) that can be neutralized by the antacid tablet. The balanced chemical equation for the neutralization reaction between HCl and magnesium hydroxide is:
2HCl + Mg(OH)2 -> 2H2O + MgCl2
From the equation, we can see that 2 moles of HCl react with 1 mole of Mg(OH)2. Therefore, the number of moles of HCl neutralized by magnesium hydroxide is:
Moles of HCl neutralized = 2 * moles of magnesium hydroxide = 2 * 0.00232 mol = 0.00464 mol
Similarly, the balanced chemical equation for the reaction between HCl and sodium bicarbonate is:
HCl + NaHCO3 -> H2O + CO2 + NaCl
From this equation, we can see that 1 mole of HCl reacts with 1 mole of NaHCO3. Therefore, the number of moles of HCl neutralized by sodium bicarbonate is:
Moles of HCl neutralized = moles of sodium bicarbonate = 0.00142 mol
Finally, let's calculate the total moles of HCl neutralized by one antacid tablet:
Total moles of HCl neutralized = moles of HCl neutralized by magnesium hydroxide + moles of HCl neutralized by sodium bicarbonate
= 0.00464 mol + 0.00142 mol
= 0.00606 mol
Given that the concentration of stomach acid (HCl) is 0.02M, we can now calculate the volume of acid that can be neutralized by one tablet using the following formula:
Volume (in liters) = Moles / Concentration
Volume of acid neutralized by one tablet = 0.00606 mol / 0.02 mol/L = 0.303 L or 303 mL
Therefore, the volume of acid that can be neutralized by one antacid tablet is 303 mL.