Answer to Question #114617 in General Chemistry for autumn funston

Question #114617
How many grams of aluminum will react with 1.12 grams of Iron II nitrate
1
Expert's answer
2020-05-08T14:11:23-0400

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


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