To balance the reaction \( \text{VCl}_2 + \text{Cl}_2 \rightarrow \text{VCl}_5 \), we can start by breaking down the components.
- We have vanadium dichloride (\( \text{VCl}_2 \)) and chlorine gas (\( \text{Cl}_2 \)) as reactants. The product is vanadium pentachloride (\( \text{VCl}_5 \)).
- To balance the equation, we can see that one molecule of \( \text{VCl}_5 \) contains one vanadium atom and five chlorine atoms. On the left side, each \( \text{VCl}_2 \) molecule contains two chlorine atoms.
Starting with: \[ \text{VCl}_2 + \text{Cl}_2 \rightarrow \text{VCl}_5 \]
We need to balance the chlorine atoms:
- On the right side, there are 5 chloride ions in \( \text{VCl}_5 \).
- To achieve this with \( \text{VCl}_2 \) and \( \text{Cl}_2 \), we can try using multiples of \( \text{VCl}_2 \).
Let's assume we have:
\[ 2 \text{VCl}_2 + 3 \text{Cl}_2 \rightarrow 2 \text{VCl}_5 \]
Now count the atoms:
- Left side: \( (2 , \text{V} + 2 \times 2 , \text{Cl}) + (3 \times 2 , \text{Cl}) = 2 , \text{V} + 4 , \text{Cl} + 6 , \text{Cl} = 2 , \text{V} + 10 , \text{Cl} \)
- Right side: \( 2 \times 1 , \text{V} + 2 \times 5 , \text{Cl} = 2 , \text{V} + 10 , \text{Cl} \)
This matches on both sides.
Thus, the balanced equation is:
\[ 2 \text{VCl}_2 + 3 \text{Cl}_2 \rightarrow 2 \text{VCl}_5 \]
So, the answer is:
Reactants: 2VCl2 and 3Cl2; Product: 2VCl5.