# Question bae7a

Apr 3, 2016

$\text{69 g}$

#### Explanation:

Start by taking a look at the balanced chemical equation for this reaction

${\text{Fe"_ 2"O"_ (3(s)) + color(red)(3)"H"_ (2(g)) -> color(blue)(2)"Fe"_ ((s)) + 3"H"_ 2"O}}_{\left(g\right)}$

Notice the mole ratios that exist between ferric oxide, ${\text{Fe"_2"O}}_{3}$, and hydrogen gas, ${\text{H}}_{2}$, and ferric oxide and iron metal, $\text{Fe}$.

The $1 : \textcolor{red}{3}$ mole ratio that exists between the two reactants tells you that the reaction will always consume three times as many moles of hydrogen gas than of ferric oxide.

On the other hand, the $1 : \textcolor{b l u e}{2}$ mole ratio that exists between ferric oxide and iron metal tells you that the reaction will theoretically produce two moles of the latter for every one mole of the former that takes part in the reaction.

At this point, your goal is to determine if you're dealing with a limiting reagent problem.

Use the molar masses of the two reactants to calculate how many moles of each you're adding

98.5 color(red)(cancel(color(black)("g"))) * ("1 mole Fe"_2"O"_3)/(159.69color(red)(cancel(color(black)("g")))) = "0.61682 moles Fe"_2"O"_3

5.0color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.0159color(red)(cancel(color(black)("g")))) = "2.4803 moles H"_2

Now, use the number of moles of ferric oxide to determine how many moles of hydrogen gas would be needed on order for all the moles of ferric oxide to take part in the reaction

0.61682color(red)(cancel(color(black)("moles Fe"_2"O"_3))) * (color(red)(3)color(white)(a)"moles H"_2)/(1color(red)(cancel(color(black)("mole Fe"_2"O"_3)))) = "1.8505 moles H"_2

As you can see, you have excess hydrogen gas present. This means that ferric oxide will act as a limiting reagent, i.e. it will be completely consumed before all the moles of hydrogen gas get the chance to react.

So, you know that ll the moles of ferric oxide take part in the reaction, which means that you can theoretically get

0.61682color(red)(cancel(color(black)("moles Fe"_2"O"_3))) * (color(blue)(2)color(white)(a)"moles Fe")/(1color(red)(cancel(color(black)("mole Fe"_2"O"_3)))) = "1.234 moles Fe"#

Finally, use iron metal's molar mass to determine how many grams would contain this many moles

$1.234 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Fe"))) * "55.845 g"/(1color(red)(cancel(color(black)("mole Fe")))) = color(green)(|bar(ul(color(white)(a/a)"69 g} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of hydrogen gas.