Limiting Reagents and Percent Yield II Answer Key (CHEM 1405)

Back to Chemistry 1405 Practice Problems

Write and balance the equations, then solve for the required quantity using dimensional analysis.

1. Magnesium iodide reacts with liquid bromine in a single replacement reaction.

MgI₂     +     Br₂          MgBr₂     +     I₂

a) Which is the excess reactant and how much excess remains from the reaction of 560 g of MgI₂ and 360 g of Br₂?

Limiting Reagent                                                                                                   Theoretical Yield
$560gMgI_2\times(\frac{1molMgI_2}{278.11gMgI_2})\times(\frac{1molI_2}{1molMgI_2})\times(\frac{253.80gI_2}{1molI_2})=$ 510g I₂

$360gBr_2\times(\frac{1molBr_2}{159.80gBr_2})\times(\frac{1molI_2}{1molBr_2})\times(\frac{253.80gI_2}{1molI_2})=$ 570g I₂

Excess Reagent Calculation:  $560gMgI_2\times(\frac{1molMgI_2}{278.11gMgI_2})\times(\frac{1molBr_2}{1molMgI_2})\times(\frac{253.80gBr_2}{1molBr_2})=$ 320g Br₂ (used)

360g Br₂ (total) - 320g Br₂ (used) = 40g Br₂ left over in excess

MgI₂ is the limiting reagent and Br₂ is the excess reagent.

 

b) What mass of I₂ can be formed from the reaction?

510g I₂  

 

 

 

 

2. Nickel (II) reacts with silver nitrate in a single replacement reaction.

Ni     +     2AgNO₃          2Ag     +     Ni(NO₃)₂

a) If 22.9 g of nickel and 112 g of silver nitrate are provided, which is limiting?

$22.9gNi\times(\frac{1molNi}{58.69gNi})\times(\frac{1molNi(NO_3)_2}{1molNi})\times(\frac{182.71gNi(NO_3)_2}{1molNi(NO_3)_2})=$ 71.3g Ni(NO₃)₂

Limiting Reagent                                                                                                                                    Theoretical Yield
$112gAgNO_3\times(\frac{1molAgNO_3}{169.88gAgNO_3})\times(\frac{1molNi(NO_3)_2}{2molAgNO_3})\times(\frac{182.71gNi(NO_3)_2}{1molNi(NO_3)_2})=$ 60.2g Ni(NO₃)₂

Silver nitrate (AgNO₃) is the limiting reagent.

 

b) What mass of nickel (II) nitrate can be produced?

60.2g Ni(NO₃)₂

 

 

 

 

3. Given the following reaction:           CS₂(g)     +     3O₂(g)          2SO₂(g)     +     CO₂(g)

a) If 1.60 mol of CS₂ reacts with 5.60 mol of oxygen gas, identify the excess reactant. How many moles of the excess reactant remain after the reaction?

Limiting Reagent                                  Theoretical Yield
$1.60molCS_2\times(\frac{2molSO_2}{1molCS_2})=$ 3.20 mol SO₂                          $5.60molO_2\times(\frac{2molSO_2}{3molO_2})=$ 3.73 mol SO₂

Excess Reactant Calculation:  $1.60molCS_2\times(\frac{3molO_2}{1molCS_2})=4.80molO_2$ (used)

5.60 mol O₂ (total) - 4.80 mol O₂ (used) = 0.80 mol O₂ left over in excess

 

b) How much SO₂ can be produced from the reaction?

3.20 mol SO₂

 

c) How much CO₂ can be produced from the reaction?

$1.60molCS_2\times(\frac{1molCO_2}{1molCS_2})=$ 1.60 mol CO₂

 

 

 

 

Balance the following equations and then solve for the required quantity using dimensional analysis.

4. Consider the following reaction:

2NO₂     +     O₃          N₂O₅     +     O₂

a) Calculate the percent yield for a reaction in which 0.38 g of NO₂ reacts with excess O₃ and 0.36 g of N₂O₅ is produced.

Limiting Reagent                                                                                                      Theoretical Yield
$0.38gNO_2\times(\frac{1molNO_2}{46.01gNO_2})\times(\frac{1molN_2O_5}{2molNO_2})\times(\frac{108.02gN_2O_5}{1molN_2O_5})=0.45gN_2O_5$

$\%yield=\frac{0.36gN_2O_5}{0.45gN_2O_5}\times100\%=$ 80.%

 

b) What mass of N₂O₅ will result from the reaction of 6.0 mol of NO₂ with excess O₃ if there is a 61.1% yield in the reaction?

Limiting Reagent                                                                          Theoretical Yield
$6.0molNO_2\times(\frac{1molN_2O_5}{2molNO_2})\times(\frac{108.02gN_2O_5}{1molN_2O_5})=320gN_2O_5$

$320gN_2O_5(theoretical)\times(\frac{61.1g N_2O_5 (actual) }{100gN_2O_5 (theoretical) })=$ 2.0 x 10² g N₂O₅

 

 

 

 

5. In the following equation:

2NaCl     +     H₂SO₄          2HCl     +     Na₂SO₄

a) What is the percent yield if 14.6 g HCl are produced from 30.0 g NaCl and 0.250 mol of H₂SO₄?

$30.0gNaCl\times(\frac{1molNaCl}{58.44gNaCl})\times(\frac{2molHCl}{2molNaCl})\times(\frac{36.46gHCl}{1molHCl})=$ 18.7 g HCl

Limiting Reagent                                                                               Theoretical Yield
$0.250molH_2SO_4\times(\frac{2molHCl}{1molH_2SO_4})\times(\frac{36.46gHCl}{1molHCl})=$ 18.2 g HCl

$\%yield=\frac{14.6gHCl}{18.2gHCl}\times100\%=$ 80.2%

 

b) If 25.6 g of NaCl reacts with excess H₂SO₄ and 12.3 g of Na₂SO₄ are produced, what is the percent yield?

$25.6gNaCl\times(\frac{1molNaCl}{58.44gNaCl})\times(\frac{1molNa_2SO_4}{2molNaCl})\times(\frac{142.04gNa_2SO_4}{1molNa_2SO_4})=$ 31.1g Na₂SO₄

 

$\%yield=\frac{12.3gNa_2SO_4}{31.1gNa_2SO_4}\times100\%=$ 39.5%

 

 

 

 

6. Assume the following hypothetical reaction takes place (already balanced):

2A     +     7B          4C     +     3D

a) Calculate percent yield when 0.0251 mol of A reacts to form 0.0349 mol of C.

Limiting Reagent                             Theoretical Yield
$0.0251molA\times(\frac{4molC}{2molA})=$ 0.0502 mol C                         $\%yield=\frac{0.0349molC}{0.0502molC}\times100\%=$ 69.5%

 

b) Calculate the percent yield when 189 mol of B reacts to form 39 mol of D.

$189molB\times(\frac{3molD}{7molB})=$ 81.0 mol D                                         $\%yield=\frac{39molD}{81.0molD}\times100\%=$ 48%