Part A ) 525mL of 1.50M NaCl is added to a beaker containing 378mL of 2.22M AgNO

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

94.5 % yield

78.8 % yield

46.6 % yield

49.4 % yield

Explanation / Answer

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?

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