# Question 9ae68

Mar 29, 2016

${\text{5.6 g Al"_2"O}}_{3}$

#### Explanation:

Your starting point here will be the balanced chemical equation for this synthesis reaction

$\textcolor{red}{4} {\text{Al"_text((s]) + 3"O"_text(2(g]) -> color(blue)(2)"Al"_2"O}}_{\textrm{3 \left(s\right]}}$

Notice that you have a $\textcolor{red}{4} : \textcolor{b l u e}{2}$ mole ratio between aluminium metal and aluminium oxide. This tells you that in theory, the reaction should produce $\textcolor{b l u e}{1}$ mole of aluminium oxide for every $\textcolor{red}{2}$ moles of aluminium metal that take part in the reaction.

The problem tells you the oxygen gas is in excess, so you an assume the all the aluminium will react.

Use the molar mass of aluminium metal to determine how many moles you get in that $\text{4.6-g}$ sample

4.6 color(red)(cancel(color(black)("g"))) * "1 mole Al"/(26.98color(red)(cancel(color(black)("g")))) = "0.1705"

So, the theoretical yield of the reaction, which tells you how much product will be produced for a 100% yield, will be equal to

0.1705color(red)(cancel(color(black)("moles Al"))) * (color(blue)(1)color(white)(a)"moles Al"_2"O"_3)/(color(red)(2)color(red)(cancel(color(black)("moles Al")))) = "0.08525 moles Al"_2"O"_3

Use aluminium oxide's molar mass to determine how many grams would contain that many moles

0.08525color(red)(cancel(color(black)("moles Al"_2"O"_3))) * "101.96 g"/(1color(red)(cancel(color(black)("mole Al"_2"O"_3)))) = "8.69 g"

So, if every mole of aluminium that takes part in the reaction ends up producing aluminium oxide, you can expect the reaction to form $\text{8.69 g}$ of product.

However, you know that only 64% of the aluminium that reacted actually produced aluminium oxide - this is the reaction's percent yield.

Percent yield is defined as

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{% yield" = "what you actually get"/"what you should theoretical get} \times 100 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Well, a percent yield of 64% essentially means that you only get $\text{64 g}$ of product for every $\text{100 g}$ that should be theoretically produced.

In this case, the actual yield of the reaction will be

64color(red)(cancel(color(black)("%"))) * overbrace("8.69 g"/(100color(red)(cancel(color(black)("%")))))^(color(purple)("theoretical yield")) = "5.562 g"

Rounded to two sig figs, the answer will be

"actual yield" = color(green)(|bar(ul(color(white)(a/a)"5.6 g Al"_2"O"_3color(white)(a/a)|)))#