Question

In an experiment 1.0912 g of ferric oxide is reacted with 0.4373 g of carbon monoxide gas according to the equation:
Fe2O3 + CO ==> Fe + CO2.

A. What is the limiting reactant?
B. What mass of the other reactant is in exess?
C. What mass of iron is produced?

Answer 1

@Gabebro13

Answer 2

Um... I don't know how to help with this... 0_0

Answer 3

Sorry...

Answer 4

First let's balance the equation:

\[Fe_{2}O_{3}+3CO \rightarrow 2Fe+3CO_{2}\]
To find the limiting reagent, we first need to find how many moles (n) of each reactant we have. To do this we divide mass by molar mass for each reactant:

\[n_{Fe_{2}O_{3}}=\frac{ 1.0912 \space g}{ 160 \space g/mol }=0.00682 \space mol\]
\[n_{CO}=\frac{0.4373 \space g}{28 \space g/mol}=0.01562 \space mol\]
From the balanced equation, we know there needs to be 3 times as much CO as Fe2O3, since it takes 3 moles of CO to react with every mole of Fe2O3. When we divide our moles of CO by 3, we get 0.00521 mol, which is how much Fe2O3 our sample of CO can react with. Since we have more Fe2O3 than that, some Fe2O3 will remain after the reaction is complete and so CO is the limiting reagent.

So, we know from above the 0.00521 mol of Fe2O3 is used up out of 0.00682 mol total. This means there will be 0.00682 - 0.00521 = 0.00161 mol of Fe2O3 remaining. Multiply by the molar mass to get the mass that remains: 0.00161 x 160 = 0.2576 g.

Now to find the mass of iron produced. From the balanced equation, we know 2 moles of Fe are produced for every 3 moles of CO used up (they react in a 2:3 ratio). That means we can set up this equation using our known moles of CO:

\[\frac{x \space mol \space Fe}{0.01562 \space mol \space CO}=\frac{2 \space mol \space Fe}{3 \space mol \space CO}\]
\[x=0.01041 \space mol\]
We know the moles of Fe now, so just multiply by the molar mass to find the mass: 0.01041 x 56 = 0.5830 g.

And that's it! If you have any questions about any of this, please let me know!