Question

When iron reacts with oxygen, it forms iron oxide, or rust. 4 Fe + 3 O2 2 Fe2O3 If 112 g of iron (Fe) combines with 24 g of oxygen (O2), how much iron oxide is formed?

Answer 1

It has to be 144 (fe) is what i calculated.

Answer 2

107 g of $Fe_2O_3$ will be produced from the given masses of both reactants.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}$    

$\text{Moles of} Fe=\frac{112g}{56g/mol}=2moles$

$\text{Moles of} O_2=\frac{24g}{32g/mol}=0.75moles$

$4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3$

According to stoichiometry :

3 moles of $O_2$ require 4 moles of $Fe$

Thus 0.75 moles of $O_2$ will require=$\frac{4}{3}\times 0.75=1mole$  of $Fe$

Thus $O_2$ is the limiting reagent as it limits the formation of product and $Fe$ is the excess reagent.

As 3 moles of $O_2$ give = 2 moles of $Fe_2O_3$

Thus 1 mole of $O_2$ give =$\frac{2}{3}\times 1=0.67moles$  of $Fe_2O_3$

Mass of $Fe_2O_3=moles\times {\text {Molar mass}}=0.67moles\times 160g/mol=107g$

Thus 107 g of $Fe_2O_3$ will be produced from the given masses of both reactants.

Learn more about stoichiometry

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