# Question 51b48

Oct 2, 2017

Here's what I got.

#### Explanation:

The underlying principle here is that the number of atoms of each element that takes part in a chemical reaction remains constant.

In other words, the total number of atoms of an element present on the reactants' side must be equal to the total number of atoms of that element present on the products' side.

In your case, the chemical reaction is said to involve atoms belonging to two types of elements, $\text{X}$ and $\text{Y}$.

On the reactants' side, you have

• three atoms of "X: " { ("2 atoms of X from X"_ 2), ("1 atom of X from XY"_2) :}
• two atoms of ${\text{Y: " "both from XY}}_{2}$

On the products' side, you have

• two atoms of ${\text{X: " "both from X"_2"Y}}_{3}$
• three atoms of ${\text{Y: " "all 3 from X"_2"Y}}_{3}$

Now, the unbalanced chemical equation looks like this

${\text{X"_ 2 + "XY"_ 2 -> "X"_ 2"Y}}_{3}$

Your goal now is to find the coefficients that can be used to balance this chemical equation.

Notice that you can get the same number of atoms of $\text{Y}$ on both sides, i.e. $6$ atoms of $\text{Y}$, if you multiply ${\text{XY}}_{2}$ by $3$ and ${\text{X"_2"Y}}_{3}$ by $2$.

This will give you

${\text{X"_ 2 + 3"XY"_ 2 -> 2"X"_ 2"Y}}_{3}$

Now focus on the atoms of $\text{X}$. You now have

overbrace("2 atoms X")^(color(blue)("from X"_2)) + overbrace("3 atoms X")^(color(blue)("from 3 XY"_2)) = "5 atoms X"

on the reactants' side and $4$ atoms of $\text{X}$--from $2 {\text{X"_2"Y}}_{3}$--on the products' side.

To balance out the atoms of $\text{X}$, you can use a fractional coefficient of $\frac{1}{2}$ in front of ${\text{X}}_{2}$. This will give you

overbrace("1 atom X")^(color(blue)("from"color(white)(.) 1/2"X"_2)) + overbrace("3 atoms X")^(color(blue)("from 3 XY"_2)) = "4 atoms X"#

on the reactants' side. This means that one version of the balanced chemical equation will look like this

$\frac{1}{2} {\text{X"_ 2 + 3"XY"_ 2 -> 2"X"_ 2"Y}}_{3}$

If you want, you can multiply all the coefficients by $2$ to get rid of the fractional coefficient.

$\left(2 \cdot \frac{1}{2}\right) {\text{X"_ 2 + (2 * 3)"XY"_ 2 -> (2 * 2)"X"_ 2"Y}}_{3}$

This will get you

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{X"_2 + 6"XY"_2 -> 4"X"_2"Y}}_{3}}}}$

Now all the atoms of $\text{X}$ and of $\text{Y}$ present on the reactants' side are also present on the products' side $\to$ mass is conserved.

Finally, the balanced chemical equation tells you that reactant $\text{B}$ and the product take part in the reaction in a $6 : 4$ mole ratio.

This means that when $2.4$ moles of $\text{B}$ react with excess $\text{A}$, the reaction will produce

$2.4 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles B"))) * "4 moles product"/(6color(red)(cancel(color(black)("moles B")))) = color(darkgreen)(ul(color(black)("1.6 moles A}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the number of moles of reactant $\text{B}$.