if 100g of iron reacts with excess oxygen, how much rust should be produced?
143,64
You have to help me by explain bruh
151.72
142.9 grams
Rust is a number of oxides of iron(III); technically, we can't calculate how much rust will form since it can take more than one form. If we call rust Fe2O3 it is done as follows.
4Fe + 3O2 --> 2Fe2O3
mols Fe = grams/atomic mass = ?
Use the coefficients in the balanced equation and convert mols Fe to mols Fe2O3.
Then g Fe2O3 = mols Fe2O3 x molar mass Fe2O3 = ? This is the theoretical yield.
% Fe2O3 = ?
By the way this is not a percentage problem.
I dnt understand the coefficient and even to convert moles please help
I don't understand everything here
143,73
OK. You know how to get mols.
mols = grams Fe/atomic mass Fe = 100 g Fe/55.85 = ?. That is estimated 1.9.
To convert mols use the coefficients in the balanced equation. To convert from mols Fe to mols Fe2O3 that's
1.9 mols Fe x (2 mol Fe2O3/4 mols Fe) = 1.9 mols x (2/4) = estimated 1.9/2 = estd 0.95 mols Fe2O3
Then grams Fe2O3 produced is
estimated 0.95 mols Fe2O3 x 159.7 = ?