To balance the chemical equation \( \text{Ti}_3\text{N} + \text{MgO} \rightarrow \text{Mg}_3\text{N}_2 + \text{Ti}_2\text{O} \), we will follow these steps:
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List the number of each type of atom in reactants and products.
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Reactants:
- Titanium (Ti): 3
- Nitrogen (N): 1
- Magnesium (Mg): 1
- Oxygen (O): 1
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Products:
- Titanium (Ti): 2
- Nitrogen (N): 2
- Magnesium (Mg): 3
- Oxygen (O): 1
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Set up the equation with coefficients:
Let’s start with: \[ a \text{ Ti}_3\text{N} + b \text{ MgO} \rightarrow c \text{ Mg}_3\text{N}_2 + d \text{ Ti}_2\text{O} \]
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Balance different elements:
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Nitrogen (N): We have 1 nitrogen atom on the left and aim for 2 on the right in \( \text{Mg}_3\text{N}_2 \). Thus, we need 2 \( \text{Ti}_3\text{N} \): \[ 2 \text{Ti}_3\text{N} + b \text{MgO} \rightarrow c \text{Mg}_3\text{N}_2 + d \text{Ti}_2\text{O} \]
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This gives us:
- Left: 2 N atoms
- Right: 2 N atoms (in \( c = 1 \))
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Titanium (Ti): With 2 \( \text{Ti}_3\text{N} \), we have 6 Ti on the left and need 6 on the right (in total) \( \rightarrow 3 \text{Ti}_2\text{O} \): \[ 2 \text{Ti}_3\text{N} + b \text{MgO} \rightarrow 1 \text{Mg}_3\text{N}_2 + 3 \text{Ti}_2\text{O} \]
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Magnesium (Mg): Now we have 3 magnesium from \( \text{Mg}_3\text{N}_2 \): \[ 2 \text{Ti}_3\text{N} + 3 \text{MgO} \rightarrow 1 \text{Mg}_3\text{N}_2 + 3 \text{Ti}_2\text{O} \]
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Final coefficients:
Putting this all together, we have the balanced equation: \[ 2 \text{Ti}_3\text{N} + 3 \text{MgO} \rightarrow 1 \text{Mg}_3\text{N}_2 + 3 \text{Ti}_2\text{O} \]
Thus:
- Blank 1 = 2
- Blank 2 = 3
- Blank 3 = 1
- Blank 4 = 3
So the final answer for the coefficients is:
Blank 1: 2
Blank 2: 3
Blank 3: 1
Blank 4: 3
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