To balance the chemical equation \( \text{Ti}_3\text{N} + \text{MgO} \rightarrow \text{Mg}_3\text{N}_2 + \text{Ti}_2\text{O} \), follow these steps:
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Write down the unbalanced equation: \[ \text{Ti}_3\text{N} + \text{MgO} \rightarrow \text{Mg}_3\text{N}_2 + \text{Ti}_2\text{O} \]
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Count the number of each type of atom on both sides:
- Reactants:
- Ti: 3
- N: 1
- Mg: 1
- O: 1
- Products:
- Ti: 2
- N: 2
- Mg: 3
- O: 1
- Reactants:
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Start balancing the most complex compound or the element that appears in the smallest quantity:
- Start with nitrogen (N). There is 1 N on the left and 2 N on the right. Therefore, place a coefficient of 2 in front of \(\text{Ti}_3\text{N}\) on the left: \[ 2 \text{Ti}_3\text{N} + \text{MgO} \rightarrow \text{Mg}_3\text{N}_2 + \text{Ti}_2\text{O} \]
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Update atom counts:
- Reactants:
- Ti: 6 (from \(2 \times \text{Ti}_3\text{N}\))
- N: 2 (from \(2 \times \text{Ti}_3\text{N}\))
- Mg: 1
- O: 1
- Products:
- Ti: 2
- N: 2
- Mg: 3
- O: 1
- Reactants:
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Now balance titanium (Ti): There are 6 Ti on the left and 2 Ti on the right. Place a coefficient of 3 in front of \(\text{Ti}_2\text{O}\): \[ 2 \text{Ti}_3\text{N} + \text{MgO} \rightarrow \text{Mg}_3\text{N}_2 + 3 \text{Ti}_2\text{O} \]
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Update atom counts:
- Reactants:
- Ti: 6
- N: 2
- Mg: 1
- O: 1
- Products:
- Ti: 6 (from \(3 \times \text{Ti}_2\text{O}\))
- N: 2
- Mg: 3
- O: 3 (from \(3 \times \text{Ti}_2\text{O}\))
- Reactants:
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Finally, balance magnesium (Mg): There are 3 Mg on the right, so place a coefficient of 3 in front of \(\text{MgO}\): \[ 2 \text{Ti}_3\text{N} + 3 \text{MgO} \rightarrow \text{Mg}_3\text{N}_2 + 3 \text{Ti}_2\text{O} \]
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Final check of atom counts:
- Reactants:
- Ti: 6
- N: 2
- Mg: 3
- O: 3
- Products:
- Ti: 6
- N: 2
- Mg: 3
- O: 3
- Reactants:
Since both sides are balanced, the final balanced equation is: \[ \boxed{2 \text{Ti}_3\text{N} + 3 \text{MgO} \rightarrow \text{Mg}_3\text{N}_2 + 3 \text{Ti}_2\text{O}} \]
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