To balance the chemical equation \( \text{NaBr} + \text{CaF}_2 \rightarrow \text{NaF} + \text{CaBr}_2 \), we will determine the coefficients that make the number of each atom on the reactant side equal to the number on the product side.
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Write down the number of atoms of each element on both sides.
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Reactants:
- Na: 1 (from NaBr)
- Br: 1 (from NaBr)
- Ca: 1 (from CaF\(_2\))
- F: 2 (from CaF\(_2\))
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Products:
- Na: 1 (from NaF)
- Br: 2 (from CaBr\(_2\))
- Ca: 1 (from CaBr\(_2\))
- F: 1 (from NaF)
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Let's balance the bromine (Br) first:
- We have 1 Br in NaBr and 2 Br in CaBr\(_2\). So we need to put a coefficient of 2 in front of NaBr: \[ 2 \text{NaBr} + \text{CaF}_2 \rightarrow \text{NaF} + \text{CaBr}_2 \]
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Now the equation looks like this:
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Reactants:
- Na: 2 (from \(2\text{NaBr}\))
- Br: 2 (from \(2\text{NaBr}\))
- Ca: 1 (from CaF\(_2\))
- F: 2 (from CaF\(_2\))
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Products:
- Na: 1 (from NaF)
- Br: 2 (from CaBr\(_2\))
- Ca: 1 (from CaBr\(_2\))
- F: 1 (from NaF)
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We still need to balance sodium (Na) next. We have 2 Na in the reactants, so we need to add a coefficient of 2 in front of NaF: \[ 2 \text{NaBr} + \text{CaF}_2 \rightarrow 2 \text{NaF} + \text{CaBr}_2 \]
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Now the final equation is:
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Reactants:
- Na: 2
- Br: 2
- Ca: 1
- F: 2
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Products:
- Na: 2
- Br: 2
- Ca: 1
- F: 2
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Both sides are now balanced. The coefficients are: \[ \text{Final balanced equation:} \quad 2 \text{NaBr} + 1 \text{CaF}_2 \rightarrow 2 \text{NaF} + 1 \text{CaBr}_2 \] or simplified: \[ 2 \text{NaBr} + \text{CaF}_2 \rightarrow 2 \text{NaF} + \text{CaBr}_2 \]
So, the coefficients are:
- 2 for NaBr
- 1 for CaF\(_2\)
- 2 for NaF
- 1 for CaBr\(_2\)