# Question #51b48

##### 1 Answer

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,

On the **reactants' side**, you have

three atomsof#"X: " { ("2 atoms of X from X"_ 2), ("1 atom of X from XY"_2) :}# two atomsof#"Y: " "both from XY"_2#

On the **products' side**, you have

two atomsof#"X: " "both from X"_2"Y"_3# three atomsof#"Y: " "all 3 from X"_2"Y"_3#

Now, the *unbalanced* chemical equation looks like this

#"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 **atoms** of

This will give you

#"X"_ 2 + 3"XY"_ 2 -> 2"X"_ 2"Y"_ 3#

Now focus on the atoms of

#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 **atoms** of

To balance out the atoms of **fractional coefficient** of

#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

#1/2"X"_ 2 + 3"XY"_ 2 -> 2"X"_ 2"Y"_ 3#

If you want, you can multiply all the coefficients by

#(2 * 1/2)"X"_ 2 + (2 * 3)"XY"_ 2 -> (2 * 2)"X"_ 2"Y"_ 3#

This will get you

#color(darkgreen)(ul(color(black)("X"_2 + 6"XY"_2 -> 4"X"_2"Y"_3)))#

Now all the atoms of **mass is conserved**.

Finally, the balanced chemical equation tells you that reactant **mole ratio**.

This means that when **moles** of

#2.4 color(red)(cancel(color(black)("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