Question #2247

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

Expert's answer

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]
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Canada #340153 on Dec 2023