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 follow these steps:
- Count the number of atoms of each element in the reactants and products.
- Adjust the coefficients to balance each element.
The unbalanced equation is:
\[ \text{Ti}_3\text{N} + \text{MgO} \rightarrow \text{Mg}_3\text{N}_2 + \text{Ti}_2\text{O} \]
Balancing process:
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On the left side, we have:
- Titanium (Ti): 3
- Nitrogen (N): 1
- Magnesium (Mg): 1
- Oxygen (O): 1
-
On the right side, we have:
- Titanium (Ti): 2
- Nitrogen (N): 2
- Magnesium (Mg): 3
- Oxygen (O): 1
Next, we proceed to balance the equation:
-
Balancing Nitrogen (N): There is 1 Nitrogen on the left and 2 on the right. To balance it, we will put a coefficient of 2 in front of \( \text{Ti}_3\text{N} \):
\[ 2 \text{Ti}_3\text{N} + \text{MgO} \rightarrow \text{Mg}_3\text{N}_2 + \text{Ti}_2\text{O} \]
Now we have:
- Left: N = 2
- Right: N = 2
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Balancing Titanium (Ti): Now we have 6 Titanium on the left (from 2 \( \text{Ti}_3\text{N} \)) and 2 on the right. To balance it, we need a coefficient of 3 in front of \( \text{Ti}_2\text{O} \):
\[ 2 \text{Ti}_3\text{N} + \text{MgO} \rightarrow \text{Mg}_3\text{N}_2 + 3 \text{Ti}_2\text{O} \]
Now we have:
- Left: Ti = 6
- Right: Ti = 6
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Balancing Magnesium (Mg): Now we have 3 Magnesium on the right, so we need a coefficient of 3 in front of \( \text{MgO} \):
\[ 2 \text{Ti}_3\text{N} + 3 \text{MgO} \rightarrow \text{Mg}_3\text{N}_2 + 3 \text{Ti}_2\text{O} \]
Now the balanced equation is:
\[ 2 \text{Ti}_3\text{N} + 3 \text{MgO} \rightarrow \text{Mg}_3\text{N}_2 + 3 \text{Ti}_2\text{O} \]
The coefficients are:
- Blank 1: 2
- Blank 2: 3
- Blank 3: 1 (which can be left out as it is implicitly understood)
- Blank 4: 3
So the answer is:
Blank 1: 2
Blank 2: 3
Blank 3: 1
Blank 4: 3