Question

acetylene can be prepared from series of reactions from calicium carbonate.the mass of 80% calicium carbonate required to prepare 2 moles of acetylene

Answer 1

Answer:

250 gm

Explanation:

the equations are

CaCO3 ⇒ CaO +CO2

CaO + 3C ⇒ CaC2+ CO

CaC2 + 2H2O ⇒ Ca(OH)2 +C2H2(ACETYLENE)

On rearranging the equations we get

2H2O + CaCO3 +3C ⇒ CO +Ca(OH)2 +C2H2 +CO2

∴ 1 mole of CaCO3 produces 1 mole of C2H2

2 moles of C2H2 means 2 moles of CaCO3

2 moles of CaCO3 weighs 200 gm of 100% pure CaCO

but here

80%(total mass of CaCO3) = 200

(80÷100) × X = 200

X = (200× 100)÷80

X = 250 gm