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Answer 1 · 2017-12-18T18:30:32

6.95g

The equation for the decomposition of Iron (III) oxide ( $Fe_2O_3$ ) is;

$2Fe_2O_3->3O_2+4Fe$

We need to know how many mols of $Fe_2O_3$ we have in 23.11g

$"number of mols" = "mass(g)"/"Relative Molecular Mass(g per mol)"$ = $"23.11"/"159.69"$ = 0.14mols

From the reaction equation, we know that 2 mols of $Fe_2O_3$ decomposes to produce 3 mols of $O_2$ .

Therefore 0.14mols of $Fe_2O_3$ will produce 0.21mols of $O_2$

Rearrange the previously used equation.

$"mass (g)" = "number of mols"*"Relative Molecular Mass(g per mol)"$ = $0.21*32$ = 6.95g