Question

Consider the following Reaction.

Mg(OH)2 (aq) + H3PO4 (l) ---> Mg3(PO4)2 (aq) + H2O (l)

A chemist allows 30.6g of Mg(OH)2 and 63.6g H3PO4 to react. When the reaction is finished, the chemist collects 34.7g Mg3(PO4)2. Determine the limiting reagent, theoretical yield, and percent yield for the reaction.

Answer 1

Hey there!:

Molar mass of Mg(OH)2 = 58.33 g/mol

number of moles Mg(OH)2 :

moles of Mg(OH)2 = 30.6 / 58.33 => 0.5246 moles

Molar mass of H3PO4 = 97.99 g/mol

number of moles H3PO4:

moles of Mg(OH)2 = 63.6 / 97.99 => 0.649 moles

Balanced chemical equation is:

3 Mg(OH)2 + 2 H3PO4 ---> Mg3(PO4)2 + 6 H2O

3 mol of Mg(OH)2 reacts with 2 mol of H3PO4 ,for 0.5246 moles of Mg(OH)2, 0.3498 moles of H3PO4 is required , but we have 0.649 moles of H3PO4, so, Mg(OH)2 is limiting reagent !

Now , we will use Mg(OH)2 in further calculation .

Molar mass of Mg3(PO4)2 = 262.87 g/mol

According to balanced equation :

mol of Mg3(PO4)2 formed = (1/3)* moles of Mg(OH)2

= (1/3)*0.5246

= 0.1749 moles of Mg3(PO4)2

use :

mass of Mg3(PO4)2 = number of mol * molar mass

= 0.1749 * 262.87

= 46 g of Mg3(PO4)2

Therefore:

% yield = actual mass * 100 / theoretical mass

% = 34.7 * 100 / 46

% = 3470 / 46

= 75.5%

Hope that helps!