Question

#2 balance this chemical equation on your own
_ rb + _ o₂ → ___ rb₂o

Explanation:

Step1: Balance oxygen atoms

First, count O atoms: reactant has 2 (from $\text{O}_2$), product has 1 (from $\text{Rb}_2\text{O}$). Multiply $\text{Rb}_2\text{O}$ by 2 to balance O:

$$\text{Rb} + \text{O}_2 ightarrow 2\text{Rb}_2\text{O}$$

Now O atoms: 2 on both sides.

Step2: Balance rubidium atoms

Count Rb atoms: product has $2 \times 2 = 4$. Multiply $\text{Rb}$ by 4 to match:

$$4\text{Rb} + \text{O}_2 ightarrow 2\text{Rb}_2\text{O}$$

Step3: Verify atom counts

Check final counts:

• Rb: 4 on reactant, $2\times2=4$ on product
• O: 2 on reactant, $2\times1=2$ on product

All atoms are balanced.

Answer:

$4\text{Rb} + 1\text{O}_2
ightarrow 2\text{Rb}_2\text{O}$
(The coefficients are 4, 1, 2 respectively)