To solve the problem of multiplying across the bottom of the unit conversion, we start by analyzing the expression provided:
[tex]\[
\begin{array}{c|c}
24 \text{ moles } C_2H_6 & 7 \text{ moles } O_2 \\
\hline
& 2 \text{ moles } C_2H_6
\end{array}
\][/tex]
1. Identify the variables:
- We have [tex]\(24\)[/tex] moles of [tex]\(C_2H_6\)[/tex] in the numerator left part.
- We need to multiply [tex]\(24\)[/tex] moles of [tex]\(C_2H_6\)[/tex] by [tex]\(7\)[/tex] moles of [tex]\(O_2\)[/tex].
2. Calculate the product of the numerator:
- [tex]\[24 \text{ moles } C_2H_6 \times 7 \text{ moles } O_2 = 168 \text{ (moles product)} \][/tex]
3. Identify the denominator:
- The denominator provided is [tex]\(2\)[/tex] moles of [tex]\(C_2H_6\)[/tex].
4. Rewriting the new fraction obtained after multiplication:
- [tex]\[
\frac{24 \text{ moles } C_2H_6 \times 7 \text{ moles } O_2}{2 \text{ moles } C_2H_6} = \frac{168 \text{ units}}{2 \text{ moles } C_2H_6}
\][/tex]
5. Result:
The product of the numerator is [tex]\(168\)[/tex], and the denominator is [tex]\(2\)[/tex], confirming our fraction:
[tex]\[
\frac{168}{2}
\][/tex]
So, the product of the denominator is [tex]\(2\)[/tex].
