How many grams of CaCO3 are needed to react with HCl to produce 80g of calcium chloride

CaCO

3

 is the reactant in excess.

The amount of  

CaCO

3

remaining is  

11.5 g

.

Explanation:

Write a balanced equation.

CaCO

3

+

2HCl

→

CaCl

2

+

CO

2

+

H

2

O

Use the balanced equation to determine the mole ratios between  

CaCO

3

and  

CaCl

2

and between  

HCl

and  

CaCl

2

and between  

CaCO

3

and  

HCl

Calcium carbonate and calcium chloride.

1 mol CaCO

3

1 mol CaCl

2

and  

1 CaCl

2

1 mol CaCO

3

Hydrochloric acid and calcium chloride.

2 mol HCl

1 CaCl

2

and  

1 mol CaCl

2

2 HCl

Calcium carbonate and hydrochloric acid.

1 mol CaCO

3

2 mol HCl

and  

2 mol HCl

1 mol CaCO

3

Determine the moles of each reactant by dividing the given masses by their molar masses.

28.0

g CaCO

3

×

1

mol CaCO

3

100.1

g CaCO

3

=

0.280 mol CaCO

3

12.0

g HCl

×

1

mol HCl

36.5

g HCl

=

0.329 mol HCl

Determine the mass of  

CaCl

2

produced by each reactant by multiplying the moles of each reactant times the mole ratios with  

CaCl

2

in the numerator. Then multiply the result by the molar mass of  

CaCl

2

(

111 g/mol

)

.

Calcium carbonate

0.280

mol CaCO

3

×

1

mol CaCl

2

1

mol CaCO

3

×

111

g CaCl

2

1

mol CaCl

2

=

31.1 g CaCl

2

Hydrochloric acid

0.329

mol HCl

×

1

mol CaCl

2

2

mol HCl

×

111

g CaCl

2

1

mol CaCl

2

=

18.3 g CaCl

2

Since hydrochloric acid yields less calcium chloride than calcium carbonate, it is the limiting reactant .

Calcium carbonate is the excess reactant .

Determine the mass of  

CaCO

3

that reacted with the limiting reactant  

HCl

.

0.329

mol HCl

×

1

mol CaCO

3

2

mol HCl

×

100.1

g CaCO

3

1

mol CaCO

3

=

16.5 g CaCO

3

.

reacted

To determine the mass of  

CaCO

3

that remains, subtract the mass that reacted from the starting mass.

28.0 g

−

16.5 g

=

11.5 g

Explanation:

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explanation: