Question

Se ard in clor 190,09 g fier cu 12% impuritati. a) Ce cantitate de clorura de fier 3 FeCl3 se formeaza? b) Ce nr de moli de clor se consuma?

Answer 1

100-12=88% pur

p=mp/mimp*100=>mp=p*mimp/100=88*190,09/100=167,2792g Fe

MFe=56g/mol

n=m/M=167,2792/56=2,9871 moli Fe

2Fe + 3Cl2 => 2FeCl3

a) 2 moli Fe..................2 moli FeCl3
2,9871 moli Fe.............x=2,9871 moli FeCl3

MFeCl3=162,5g/mol

m FeCl3=2,9871*162,5=485,4g FeCl3

b) 2 moli Fe..........3 moli Cl2
2,9871 moli Fe......x moli Cl2

x=3*2,9871/2 =