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]
Our fields of expertise
Submit Your Assignment. We Can Do It.
LATEST TUTORIALS
APPROVED BY CLIENTS
"assignmentexpert.com" is professional group of people in Math subjects! They did assignments in very high level of mathematical modelling in the best quality. Thanks a lot
Canada #339914 on Dec 2023