The active ingredients of an antacid tablet contained only magnesium hydroxide and aluminium hydroxide. Complete neutralization of a sample of the active ingredients required 48.5 ML of 0.187 M hydrochloric acid. The chloride salts from this neutralization were obtained by evaporation of the filtrate from titration; they weighed 0.4200g.what was the percentage by mass of magnesium hydroxide in the active ingredients of the antacid tablet?
To find the percentage by mass of magnesium hydroxide in the antacid tablet, we need to calculate the moles of magnesium hydroxide reacted during the neutralization and then divide it by the total mass of the antacid tablet.
Step 1: Calculate the moles of hydrochloric acid used in the neutralization.
Molarity of hydrochloric acid (HCl) = 0.187 M
Volume of hydrochloric acid used = 48.5 mL = 0.0485 L
Moles of HCl = Molarity x Volume = 0.187 M x 0.0485 L = 0.0090595 mol
Step 2: Convert the moles of HCl reacted to moles of magnesium hydroxide.
In the balanced equation, the ratio between HCl and magnesium hydroxide is 2:2, which means 2 moles of HCl react with 2 moles of magnesium hydroxide.
Moles of magnesium hydroxide = Moles of HCl = 0.0090595 mol
Step 3: Calculate the molar mass of magnesium hydroxide.
Molar mass of magnesium (Mg) = 24.31 g/mol
Molar mass of oxygen (O) = 16.00 g/mol
Molar mass of hydrogen (H) = 1.01 g/mol
Molar mass of magnesium hydroxide (Mg(OH)2) = (24.31 g/mol) + 2[(16.00 g/mol) + (1.01 g/mol)]
= 24.31 g/mol + 2(17.01 g/mol)
= 58.33 g/mol
Step 4: Calculate the mass of magnesium hydroxide in the antacid tablet.
Mass of magnesium hydroxide = Moles of magnesium hydroxide x Molar mass of magnesium hydroxide
= 0.0090595 mol x 58.33 g/mol
= 0.5260 g
Step 5: Calculate the percentage by mass of magnesium hydroxide.
Percentage by mass of magnesium hydroxide = (Mass of magnesium hydroxide / Total mass of antacid tablet) x 100%
= (0.5260 g / 0.4200 g) x 100%
= 125.2%
Therefore, the percentage by mass of magnesium hydroxide in the active ingredients of the antacid tablet is 125.2%.
To find the percentage by mass of magnesium hydroxide in the active ingredients of the antacid tablet, we need to make use of the information given.
Step 1: Calculate the number of moles of hydrochloric acid used in the neutralization.
Moles of HCl = Volume of HCl (in L) x Molarity of HCl
Given:
Volume of HCl = 48.5 mL = 48.5/1000 L = 0.0485 L
Molarity of HCl = 0.187 M
Moles of HCl = 0.0485 L x 0.187 M = 0.0090695 mol
Step 2: Determine the molar ratio between magnesium hydroxide (Mg(OH)2) and hydrochloric acid (HCl) in the neutralization reaction.
From the balanced chemical equation:
Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
The molar ratio between Mg(OH)2 and HCl is 1:2. This means that 1 mole of Mg(OH)2 reacts with 2 moles of HCl.
Step 3: Calculate the number of moles of Mg(OH)2 used in the neutralization.
Moles of Mg(OH)2 = (0.0090695 mol HCl) / 2 = 0.00453475 mol Mg(OH)2
Step 4: Determine the molar mass of Mg(OH)2.
Molar mass of Mg(OH)2 = (24.31 g/mol) + 2 * ((1.01 g/mol) + (16.00 g/mol)) = 58.33 g/mol
Step 5: Calculate the mass of Mg(OH)2 used in the neutralization.
Mass of Mg(OH)2 = Moles of Mg(OH)2 x Molar mass of Mg(OH)2
Mass of Mg(OH)2 = 0.00453475 mol x 58.33 g/mol = 0.2647 g
Step 6: Calculate the percentage by mass of Mg(OH)2 in the antacid tablet.
Percentage by mass = (Mass of Mg(OH)2 / Total mass of active ingredients) x 100
Percentage by mass = (0.2647 g / 0.4200 g) x 100 = 63.02%
Therefore, the percentage by mass of magnesium hydroxide in the active ingredients of the antacid tablet is approximately 63.02%.
This is a two unknowns/two equation problem. The equations are solved simultaneously. Difficult to explain on a screen but I can show it in parts and leave most of the work to you. I should note that there is a piece of information missing; i.e., you don't have the mass of the sample taken and you need to do %. The chemical equations are these.
Mg(OH)2 + 2HCl ==> MgCl2 + 2H2O
Al(OH)3 + 3HCl ==> AlCl3 + 3H2O
Let X = mass Mg(OH)2
and Y = mass Al(OH)3
-------------------------
I will let mm stand for molar mass.
Equation 1 is obtained by converting to mols HCl used in terms of X and Y. Here it is.
[2X/mm Mg(OH)2] + [3Y/mm Al(OH)3] = M HCl x L HCl
Equation 2 is obtained by converting to g of salts (MgCl2 and AlCl3) in terms of X and Y. That second equation is
[X(mm MgCl2/mm Mg(OH)2)] + [Y(mm AlCl3/mm Al(OH)3] = 0.4200 g
Solve those two equations simultaneously for X and Y which gives you grams Mg(OH)2 and grams Al(OH)3.
The % Mg(OH)2 = (X/mass sample)*100 and mass sample is not given.
%Al(OH)3 is not asked but can be done the same way.
%Al(OH)3 = (Y/mass sample)*100 -= ? and note again that mass of the sample is not given.
equation 1 I wrote for you is not right.
That should be mols HCl used for Mg(OH)2 + mols HCl used for Al(OH)3 = total mols HCl = M x L = ?
Equation 1 should read,
(X/2*mm Mg(OH)2) + (Y/3*mm Al(OH)3) = M HCl x L HCl
Sorry about that.