To balance the equation \( \text{AI} + \text{F}_2 \rightarrow \text{AIF}_3 \), we need to ensure that the number of each type of atom is the same on both sides.
Starting with the unbalanced equation:
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On the left side, we have:
- 1 Aluminum (AI)
- 2 Fluorine (F from \( F_2 \))
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On the right side, we have:
- 1 Aluminum (from AIF3)
- 3 Fluorine (from AIF3)
To balance the equation:
- We keep the 1 Aluminum atom as is on both sides.
- Since we have 3 Fluorine atoms in AIF3 and only 2 Fluorine atoms from F2, we need to adjust the coefficients to balance the fluorine atoms.
By putting a coefficient of 3 in front of AIF3 and a coefficient of 1.5 in front of F2 (to get 3 Fluorine atoms), we modify our equation as follows:
\[ \text{AI} + 1.5 , \text{F}_2 \rightarrow 3 , \text{AIF}_3 \]
However, since we usually prefer whole number coefficients, we can multiply the entire equation by 2:
\[ 2 , \text{AI} + 3 , \text{F}_2 \rightarrow 2 , \text{AIF}_3 \]
Thus, the balanced equation is:
\[ 2 , \text{AI} + 3 , \text{F}_2 \rightarrow 2 , \text{AIF}_3 \]
The coefficient that would go in front of AIF3 is 2.
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