Question

consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of cao and 13.8 g of co 2 to react. when the reaction is finished, the chemist collects 19.4 g of caco 3 . determine the limiting reactant, theoretical yield, and percent yield for the reaction.

Answer 1

CaO is the limiting reagent

Theoritical yield = 25.71 g

% Yield = 75.44%

Step-by-step explanation:

1 mole = Molar mass of the substance

Molar Mass of CaO = 56 g/mol

Molar Mass of CaCO3 = 100 g/mol

Molar mass of CO2 = 44 g/mol

The balanced Equation is :

`CaO + CO_(2)\rightarrow CaCO_(3)`

1 mole of CaO reacts with = 1 mole of CO2

56 g of CaO reacts with = 44 g of CO2

1 g of CaO reacts with =

`(44)/(56)`

= 0.785 g of CO2

So,

14.4 g of CaO must react with = (14.4 x 0.785) g of CO2

= 11.31 g of CO2

Needed = 11.31 g

Available CO2 = 13.8 g (given)

So CO2 is in excess , hence CaO is the limiting reagent and product will produce from 14.4 g of CaO

`CaO + CO_(2)\rightarrow CaCO_(3)`

1 mole of CaO will produce 1 mole pf CaCO3

56 g of CaO produce = 100 g of CaCO3

1 g of CaO produce =

`(100)/(56)`

= 1.785 g of CaCO3

14.4 g of CaO will produce = (1.785 x 14.4) g of CaCO3

= 25.71 g of CaCO3

Theoritical Yield of CaCO3 = 25.71 g

Actual yield = 19.4 g

Percent Yield =

`(Actual\ yield)/(Theoritical\ yield)* 100`

`(19.4)/(25.71)* 100`

= 75.44 %