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
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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.
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?
2 answers
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.
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.
1 answer