Answer to Question #2247 in Inorganic Chemistry for Amna
How many grams of iron(III) nitrate could be formed from the reaction of 2.535 g of iron with excess HNO3?
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]