An antacid tablet containing magnesium hydroxide instead of calcium carbonate was dissolved in 100.0 mL of 0.2893 M HCL. 10 mL of this solutions was titrated to endpoint with 25.20 mL of 0.1007 M NaOH. How much magnesium hydroxide was in the antacid tablet(in mg)?
Any help is appreciated
Mg(OH)2 + 2HCl ==> MgCl2 + 2H2O
millimols HCl added = mL x M = 100 x 0.2893M = 28.93 total acid added.
mmols HCl not used = 25.20 mL x 0.1007 = 2.538 for a 10 mL portion of the above acid; therefore, if 100 mL had been titrated you would have used 25.38 mmols NaOH which corresponds to 25.38 mmols acid not used. So how much acid was used.
begin = 28.93
-not used = 25.38
----------------
used = 3.55 mmols.
3.55 mmols acid used = 1/2 that for mmols Mg(OH)2 in the sample. Convert that to grams if you want to know grams Mg(OH)2 in one tablet.
To find the amount of magnesium hydroxide in the antacid tablet, we can use the stoichiometry of the reaction between magnesium hydroxide (Mg(OH)2) and hydrochloric acid (HCl), as well as the titration with sodium hydroxide (NaOH).
First, let's write the balanced equation for the reaction between Mg(OH)2 and HCl:
Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
From the balanced equation, we can see that 1 mole of Mg(OH)2 reacts with 2 moles of HCl.
Next, let's calculate the number of moles of HCl that reacted in the titration using the given concentration and volume:
Volume of HCl used in titration = 10 mL = 0.01 L
Moles of HCl used in titration = concentration * volume
= 0.2893 M * 0.01 L
Since the stoichiometric ratio between HCl and Mg(OH)2 is 2:1, the number of moles of Mg(OH)2 reacted with the HCl is half of the moles of HCl used in titration.
Moles of Mg(OH)2 = (Moles of HCl used in titration) / 2
Now, let's calculate the moles of Mg(OH)2:
Moles of HCl used in titration = 0.2893 M * 0.01 L
= 0.002893 moles
Moles of Mg(OH)2 = 0.002893 moles / 2
To find the mass of Mg(OH)2, we need to multiply the moles by the molar mass of Mg(OH)2:
Molar mass of Mg(OH)2 = (24.31 g/mol) + (2 * 1.01 g/mol) + (2 * 16.00 g/mol)
= 58.33 g/mol
Mass of Mg(OH)2 = Moles of Mg(OH)2 * Molar mass of Mg(OH)2
Finally, let's calculate the mass of Mg(OH)2:
Mass of Mg(OH)2 = 0.002893 moles * 58.33 g/mol
Therefore, the amount of magnesium hydroxide in the antacid tablet is approximately (mass) mg.
To find out how much magnesium hydroxide was in the antacid tablet, we need to determine the number of moles of HCl that reacted with the antacid and then use stoichiometry to calculate the amount of magnesium hydroxide present.
1. Calculate the number of moles of HCl used:
Moles of HCl = molarity of HCl * volume of HCl (in liters)
= 0.2893 M * (10.0 mL / 1000 mL/ L)
= 0.002893 mol
2. Determine the number of moles of NaOH used:
Moles of NaOH = molarity of NaOH * volume of NaOH (in liters)
= 0.1007 M * (25.20 mL / 1000 mL/ L)
= 0.002541 mol
3. Since HCl and NaOH react in a 1:1 ratio, the number of moles of HCl reacted with the antacid is equal to the number of moles of NaOH:
Moles of HCl reacted with antacid = Moles of NaOH
= 0.002541 mol
4. Next, we need to determine the molar ratio between HCl and magnesium hydroxide (Mg(OH)2):
From the balanced chemical equation:
2 HCl + Mg(OH)2 → 2 H2O + MgCl2
The ratio between HCl and magnesium hydroxide is 2:1, which means that 2 moles of HCl react with 1 mole of magnesium hydroxide.
5. Calculate the number of moles of magnesium hydroxide:
Moles of Mg(OH)2 = 0.5 * Moles of HCl reacted with antacid
= 0.5 * 0.002541 mol
= 0.0012705 mol
6. Finally, calculate the mass of magnesium hydroxide:
Mass of Mg(OH)2 = moles of Mg(OH)2 * molar mass of Mg(OH)2
The molar mass of Mg(OH)2 can be found by adding the atomic masses of magnesium (24.31 g/mol), oxygen (16.00 g/mol), and hydrogen (1.01 g/mol) together:
Molar mass of Mg(OH)2 = (24.31 g/mol) + 2 * (1.01 g/mol) + 2 * (16.00 g/mol)
= 58.33 g/mol
Mass of Mg(OH)2 = 0.0012705 mol * 58.33 g/mol
= 0.07405 g
Finally, convert the mass to milligrams:
Mass of Mg(OH)2 = 0.07405 g * 1000 mg/g
= 74.05 mg
Therefore, there is approximately 74.05 mg of magnesium hydroxide in the antacid tablet.