Question

Part A) 525mL of 1.50M NaCl is added to a beaker containing 378mL of 2.22M AgNO3, resulting in the formation of AgCl precipitate. What is the limiting reactant?

NaCl

AgNO3

AgCl

NaNO3

Part B) How many grams of K₂SO₄ are in 250mL of 0.11 M K₂SO₄ solution?

___ g K₂SO₄

Part C) The reaction of magnesium nitride and water is shown below:

Mg₃N₂ + 3H₂O --> 2NH₃ + 3MgO

During a lab experiment, 3.82 g magnesium nitride and 7.73 g water are placed in a reaction flask.

At the end of the experiment, the amount of magnesium oxide produced is 3.60 g.

What is the percent yield of the reaction?

		94.5 % yield
		78.8 % yield
		46.6 % yield
		49.4 % yield

Answer 1

Part A :

answer : NaCl

moles of NaCl = 525 x 1.5 / 1000 = 0.7875

moles of AgNO3 = 378 x 2.22 / 1000 = 0.839

NaCl + AgNO3   ----------------> AgCl + NaNO3

1                1      

0.7875       0.839

here the limiting reagent is NaCl.

Part B) answer : 4.79 g

moles of K2SO4 = 0.11 x 250 / 1000 = 0.0275

mass of K2SO4 = moles x molar mass = 0.0275 x 174.3 = 4.79 g

mass of K2SO4 = 4.79 g

Part C) answer : 78.8 %

Mg₃N₂   +    3H₂O ---------------->    2NH₃    +   3MgO

100.9 g       54 g                              34 g         120 g

3.82 g    7.73 g                                              ??

here limiting reagent is Mg3N2. so MgO formed based on that.

100.9 g of Mg3N2 ---------------> 120 g of MgO

3.82 g of Mg3N2 ----------------> ?? MgO

theoretical yield of MgO = 3.82 x 120 / 100.9 = 4.54 g

actual yield = 3.60 g.

percent yield = (actual / theoretical ) x 100

                     = (3.60 / 4.54 ) x 100

                    = 78.8 %

What is the percent yield of the reaction?