Question

The reaction below consistently yields 82% of the theoretical value. If you need to make 10 g of Mg(OH)2 How much Mg3N2 (g) should you use (assume that you will use H2O in excess).
Mg3N2(s) + H2O(l) → Mg(OH)2(s) + NH3(g)
Help me please!

Answer 1

If excess $H_2O$ is used as stated in the illustration the amount of $Mg_3N_2$ that would be required would be 7.57 grams

Stoichiometric calculation

From the equation of the reaction:

$Mg_3N_2(s) + 6H_2O(l) --- > 3Mg(OH)_2(s) + 2NH_3(g)$

Since the yield is 80% of the theoretical value, in order to make 10g, the theoretical yield would be 12.5 g.

Mole ratio of $Mg_3N_2$ and $Mg(OH)_2$ = 3:1

Mole of 12.5 g  $Mg(OH)_2$ = 12.5/53.32 = 0.225moles

Equivalent mole of $Mg_3N_2$  = 0.124/3 = 0.075 moles

Mass of 0.0041 moles  $Mg_3N_2$ = 0.075 x 100.95 = 7.57 grams

More on stoichiometric calculations can be found here: brainly.com/question/8062886