Practice Stoichiometry Problems: (Show unit-factor setup for each problem) Consi

Question

Practice Stoichiometry Problems: (Show unit-factor setup for each problem) Consider the following reaction between solid potassium carbonate and hydrochloric acid: If you wanted to produce 2.00 g of KCl a. How many grams of K,CO, should you use along with excess HCI? b. How many grams of CO, would be produced as a byproduct? How many grams of water would be produced as a byproduct? c. d. How many grams of KCI would you produce if you started with 4.00g of K,CO? e. If you actually produced 4.24 g of KCl, what was your percentage yield?

Explanation / Answer

a)

mol of KCl = mass of KCL /( MW of KCl ) = 2/74.5513 = 0.026827

ratio is 1/2 so,

1/2 *0.026827 mol of K2CO3 = 0.0134135 mol of K2CO3

mass = mol*MW = 0.0134135*138.2055 = 1.853 g of K2CO3

b)

if 0.026827 mol of KCl were produced, then

mol of CO2 = 1/2*mol of KCl = 1/2*0.026827 = 0.0134135 mol of CO2

mass = mol*MW = 0.0134135*44 = 0.590194 g of CO2

c)

1 mol of water = 1 mol of CO2

0.0134135 mol of CO2 = 0.0134135 mol of water

mass of water --> mol*MW = (0.0134135 )(18) = 0.241443 g of water

d)

mass of KCl given -- > 4 g of K2CO3

mol of K2CO3 = mass/MW = 4/138.2055 = 0.0289 mol of K2CO3

2 mol of KCl -->1 mol of K2CO3

mol of K2CO3 = 2*0.0289 = 0.0578 mol of KCl

mass = mol*MW = 0.0578*74.5513 = 4.309 g of KCl

e)

m = 4.24 g of KCl ar produced

% yield = real / theoretical * 100% = 4.24/4.309 *100 = 98.3987 %

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