Answer to Question #2247 in Inorganic Chemistry for Amna

Question

How many grams of iron(III) nitrate could be formed from the reaction of 2.535 g of iron with excess HNO3?

Answer 1

Fe + 4HNO₃ = Fe(NO₃)₃ + NO + 2H₂O
M(Fe) = 56
M(Fe(NO₃)₃) = 242
m(Fe(NO₃)₃) = ν M(Fe(NO₃₎₃) = 2.525[g]/56[g/mol] * 242[g/mol] = 11.363 [g]