Solution:
2Al+3Fe(NO3)2=2Al(NO3)3+3Fe
2 mol - 3 mol
"n(Fe(NO_{3}) _{2} ) =\\frac{m(Fe(NO_{3}) _{2} ) }{M(Fe(NO_{3}) _{2} ) }\n \\\\M(Fe(NO_{3}) _{2} )=56+(14+16*3)*2=180 (g\/mol) \\\\ \\\\ \nn(Fe(NO_{3}) _{2} ) \\frac{1.12 }{180 } =0.006(mol) \\\\"
x mol - 0.006 mol
2Al+3Fe(NO3)2=2Al(NO3)3+3Fe
2 mol - 3 mol
"x=n(Al)= \\frac{2*0.006}{3}=0.004 (mol) \\\\\nm(Al)=n(Al)*M(Al)\\\\"
"M(Al)=27 (g\/mol)\\\\ m(Al)=0.004*27=0.11(g)\\\\"
Answer : m(Al)=0.11 g
Comments
Leave a comment