Question

chalk is almost pure calcium carbonate if 10 gram of chalk is reacted with an excess of dilute hydrochloric acid 2128 litres of carbon dioxide gas is collected at STP. Determine the purity of calcium carbonate

Answer 1

CaCO is3

+2HCl→CaCl

+H2

O+CO

100 g 44 g

100 g 44 g1 mole 1 mole

100 g 44 g1 mole 1 mole22.4 L 22.4 L

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2Hence, 2.5 g CaCO

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2Hence, 2.5 g CaCO 3

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2Hence, 2.5 g CaCO 3

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2Hence, 2.5 g CaCO 3 will form =

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2Hence, 2.5 g CaCO 3 will form = 100

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2Hence, 2.5 g CaCO 3 will form = 1002.5×22.4

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2Hence, 2.5 g CaCO 3 will form = 1002.5×22.4 =

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2Hence, 2.5 g CaCO 3 will form = 1002.5×22.4 = 100

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2Hence, 2.5 g CaCO 3 will form = 1002.5×22.4 = 10056

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2Hence, 2.5 g CaCO 3 will form = 1002.5×22.4 = 10056

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2Hence, 2.5 g CaCO 3 will form = 1002.5×22.4 = 10056 L of CO

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2Hence, 2.5 g CaCO 3 will form = 1002.5×22.4 = 10056 L of CO 2

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2Hence, 2.5 g CaCO 3 will form = 1002.5×22.4 = 10056 L of CO 2 =0.56 L of CO

100 g 44 g1 mole 1 mole22.4 L 22.4 LWe know that 1 mole of any gas at 0 o C and 1 atm pressure occupies 22.4 L volume.So, 100 g CaCO 3 forms 22.4 L of CO 2Hence, 2.5 g CaCO 3 will form = 1002.5×22.4 = 10056 L of CO 2 =0.56 L of CO Hence, the correct option is D