Question

Consider the following 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!