Question

How many grams of aluminum will react with 1.12 grams of Iron II nitrate

Answer 1

Solution:

2Al+3Fe(NO₃)₂=2Al(NO₃)₃+3Fe

2 mol - 3 mol

$n(Fe(NO_{3}) _{2} ) =\frac{m(Fe(NO_{3}) _{2} ) }{M(Fe(NO_{3}) _{2} ) } \\M(Fe(NO_{3}) _{2} )=56+(14+16*3)*2=180 (g/mol) \\ \\ n(Fe(NO_{3}) _{2} ) \frac{1.12 }{180 } =0.006(mol) \\$

x mol - 0.006 mol

2Al+3Fe(NO3)2=2Al(NO3)3+3Fe

2 mol - 3 mol

$x=n(Al)= \frac{2*0.006}{3}=0.004 (mol) \\ m(Al)=n(Al)*M(Al)\\$

$M(Al)=27 (g/mol)\\ m(Al)=0.004*27=0.11(g)\\$

Answer : m(Al)=0.11 g