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with the balanced equation : 4Fe + 3O2 = 2Fe2O3How many grams of oxygen is needed to react with 125 grams of Fe?

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Step 1 - Reading and understanding the chemical equation

The provided chemical equation is:


4Fe+3O_2\rightarrow2Fe_2O_3

We can read this equation as follows:

4 moles of Fe react with 3 moles of O2 thus producing 2 moles of Fe2O3

As the exercise is specifically asking about the proportion between the reactants, we can further simplify this statement to:

4 moles of Fe react with 3 moles of O2

Step 2 - Converting the moles proportion to grams proportion

We can easily convert between moles and grams by using the molar mass of each substance (56 g/mol for Fe, 32 g/mol for O2):


\begin{gathered} Fe\rightarrow4\text{ moles x 56 g/mol = 224g} \\ O_2\rightarrow\text{ 3 moles x 32 g/mol = 96g} \end{gathered}

We can rewrite the proportion thus as:

224 of Fe react with 96g of O2

Step 3 - Calculating the needed amount of O2

Now we just have to set a proportion. We know 125g of Fe were used, so:


\begin{gathered} 224g\text{ of Fe produce ---- 96g of O2} \\ 125\text{ of Fe would produce --- x g of O2} \\ \\ x=(96*125)/(224)=53.6g\text{ of O2} \end{gathered}

Answer: 53.6g of O2

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