When the following equations are balanced using the smallest possible integers, what is the number in front of the underlined substance in each case?
-Li(s) + O₂(g) $\rightarrow$ Li₂O(s)

Question

When the following equations are balanced using the smallest possible integers, what is the number in front of the underlined substance in each case?
-Li(s) + O₂(g)

\rightarrow

Li₂O(s)

Quizplus · Accepted Answer

Li(s) + O₂(g) F1F1F1S1F1F1F10 Li₂O(s) The equation needs to be balanced by finding the smallest whole number coefficients that allow for the same number and types of atoms on both sides of the equation. There are 2 Li atoms and 2 F atoms on each side, so those are already balanced. However, there is only 1 O atom on the left side and 13 O atoms on the right side. This can be balanced by placing a coefficient of 13 in front of Li(s) and O₂(g) and a coefficient of 2 in front of F1F1F1S1. This gives the balanced equation: 13Li(s) + O₂(g) + 2F1F1F1S1 -> F1F1F10Li₂O(s) Therefore, the number in front of the underlined substance (Li₂O) is 10, and the correct answer is B.

When the Following Equations Are Balanced Using the Smallest Possible →\rightarrow→

When the Following Equations Are Balanced Using the Smallest Possible $\rightarrow$