Question

18.4 of mixture of caco3 and mgco3 on heating gives 4.0g of mgo the volume of co2 produced at stp in this process is

Answer 1

MgCO3+CaCO3 ------------->MgO+CaO+2CO2
number of mole of CO2 = 2*number of mole of MgO
 no.of mole of CO2 = 2*4/40 = 1/5 mole
volume of 1 mole gas at stp = 22.4L
volume of CO2 = 22.4*1/5 = 4.48L

Answer 2

Answer:4.48 litres

Explanation:

MgCO3 ------------->MgO+CO2

CaCO3 gives on heating CaO+CO2

number of mole of CO2 = 2*number of mole of MgO

no.of mole of CO2 = 2*4/40 = 1/5 mole

volume of 1 mole gas at stp = 22.4L

volume of CO2 = 22.4*1/5 = 4.48L