Question

1. __ secl₆ + o₂ → seo₂ + __ cl₂

Explanation:

Step1: Balance Se

There is 1 Se in \( \text{SeCl}_6 \) and 1 Se in \( \text{SeO}_2 \), so the coefficient for \( \text{SeCl}_6 \) and \( \text{SeO}_2 \) related to Se is 1 for now. Let's set the coefficient of \( \text{SeCl}_6 \) as \( x = 1 \), then the coefficient of \( \text{SeO}_2 \) should also be 1 to balance Se.

Step2: Balance Cl

In \( \text{SeCl}_6 \), there are 6 Cl atoms. In \( \text{Cl}_2 \), each molecule has 2 Cl atoms. So the number of \( \text{Cl}_2 \) molecules needed is \( \frac{6}{2}=3 \). So the coefficient of \( \text{Cl}_2 \) is 3.

Step3: Balance O

In \( \text{SeO}_2 \), there are 2 O atoms. In \( \text{O}_2 \), each molecule has 2 O atoms. So the number of \( \text{O}_2 \) molecules needed is \( \frac{2}{2}=1 \). So the coefficient of \( \text{O}_2 \) is 1.

Now we check the equation: \( 1\text{SeCl}_6 + 1\text{O}_2
ightarrow 1\text{SeO}_2 + 3\text{Cl}_2 \)

• Se: 1 on left, 1 on right.
• Cl: 6 on left (\( 1\times6 \)), 6 on right (\( 3\times2 \)).
• O: 2 on left (\( 1\times2 \)), 2 on right (\( 1\times2 \)).

It is balanced.

Answer:

1, 1, 1, 3 (The coefficients for \( \text{SeCl}_6 \), \( \text{O}_2 \), \( \text{SeO}_2 \), \( \text{Cl}_2 \) respectively)