Question

please help me out
only correct answer ​

Answer 1

Answer:

89.6L or 4mol

Explanation:

on balancing the reaction

2Al2O3+3C=====>3CO2+4Al

volume of CO2 given=134.4L

at stp ,p=1atm,T=273k

by PV=nRT

R=0.0821 L atm / K mol

1×134.4=n×0.0821×273

134.4=22.4n

n=134.4/22.4=6mol of CO2

now by stoichiometry

2Al2O3=====>3CO2

xAl2CO3======>6CO2

x=6×2/3=4mol Al2CO3

so now we need to find the volume of Al2O3

the reaction is at STP

so again by

PV=nRT

1×V=4×0.0821×273

V=89.6L