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Sagot :
Absolutely, let's go through the process of balancing the chemical equation step-by-step.
The given unbalanced equation is:
[tex]\[ \text{Li}_{(s)} + \text{N}_2{(g)} \rightarrow \text{Li}_3\text{N}_{(s)} \][/tex]
To balance the equation, we need to ensure that the number of atoms for each element is the same on both the reactant and product sides.
1. Start by placing coefficients for each compound:
[tex]\[ x \text{Li}_{(s)} + y \text{N}_2{(g)} \rightarrow z \text{Li}_3\text{N}_{(s)} \][/tex]
2. Count the atoms for each element in the unbalanced equation:
- Li (Lithium):
- Reactants: [tex]\(x\)[/tex] atoms of Li
- Products: [tex]\(3z\)[/tex] atoms of Li (since each [tex]\(\text{Li}_3\text{N}\)[/tex] molecule contains 3 Li atoms)
- N (Nitrogen):
- Reactants: [tex]\(2y\)[/tex] atoms of N
- Products: [tex]\(z\)[/tex] atoms of N (since each [tex]\(\text{Li}_3\text{N}\)[/tex] molecule contains 1 N atom)
3. Balance Nitrogen (N) first:
The reactant side has [tex]\(N_2\)[/tex], so it has 2 nitrogen atoms per molecule. To get 2 nitrogen atoms on the product side, we need 2 [tex]\(\text{Li}_3\text{N}\)[/tex] molecules:
[tex]\[ \text{Li}_{(s)} + \text{N}_2{(g)} \rightarrow 2 \text{Li}_3\text{N}_{(s)} \][/tex]
4. Update the count for each element with the new coefficient:
- Li (Lithium):
- Reactants: [tex]\(x\)[/tex] atoms of Li
- Products: [tex]\(3 \times 2 = 6\)[/tex] atoms of Li
- N (Nitrogen):
- Reactants: [tex]\(2\)[/tex] atoms of N
- Products: [tex]\(2\)[/tex] atoms of N
5. Balance Lithium (Li) next:
Since we need 6 lithium atoms on the product side, we should have 6 lithium atoms on the reactant side:
[tex]\[ 6 \text{Li}_{(s)} + \text{N}_2{(g)} \rightarrow 2 \text{Li}_3\text{N}_{(s)} \][/tex]
6. Confirm the balance:
- Li (Lithium):
- Reactants: [tex]\(6\)[/tex] atoms of Li
- Products: [tex]\(3 \times 2 = 6\)[/tex] atoms of Li
- N (Nitrogen):
- Reactants: [tex]\(2\)[/tex] atoms of N
- Products: [tex]\(2\)[/tex] atoms of N
Both lithium and nitrogen are balanced with 6 lithium atoms and 2 nitrogen atoms on both sides of the equation.
Balanced equation:
[tex]\[ 6 \text{Li}_{(s)} + \text{N}_2{(g)} \rightarrow 2 \text{Li}_3\text{N}_{(s)} \][/tex]
Therefore, the coefficients are:
- Lithium ([tex]\(\text{Li}\)[/tex]): 6
- Nitrogen ([tex]\(\text{N}_2\)[/tex]): 1
- Lithium Nitride ([tex]\(\text{Li}_3\text{N}\)[/tex]): 2
This completes the balance of the chemical equation.
The given unbalanced equation is:
[tex]\[ \text{Li}_{(s)} + \text{N}_2{(g)} \rightarrow \text{Li}_3\text{N}_{(s)} \][/tex]
To balance the equation, we need to ensure that the number of atoms for each element is the same on both the reactant and product sides.
1. Start by placing coefficients for each compound:
[tex]\[ x \text{Li}_{(s)} + y \text{N}_2{(g)} \rightarrow z \text{Li}_3\text{N}_{(s)} \][/tex]
2. Count the atoms for each element in the unbalanced equation:
- Li (Lithium):
- Reactants: [tex]\(x\)[/tex] atoms of Li
- Products: [tex]\(3z\)[/tex] atoms of Li (since each [tex]\(\text{Li}_3\text{N}\)[/tex] molecule contains 3 Li atoms)
- N (Nitrogen):
- Reactants: [tex]\(2y\)[/tex] atoms of N
- Products: [tex]\(z\)[/tex] atoms of N (since each [tex]\(\text{Li}_3\text{N}\)[/tex] molecule contains 1 N atom)
3. Balance Nitrogen (N) first:
The reactant side has [tex]\(N_2\)[/tex], so it has 2 nitrogen atoms per molecule. To get 2 nitrogen atoms on the product side, we need 2 [tex]\(\text{Li}_3\text{N}\)[/tex] molecules:
[tex]\[ \text{Li}_{(s)} + \text{N}_2{(g)} \rightarrow 2 \text{Li}_3\text{N}_{(s)} \][/tex]
4. Update the count for each element with the new coefficient:
- Li (Lithium):
- Reactants: [tex]\(x\)[/tex] atoms of Li
- Products: [tex]\(3 \times 2 = 6\)[/tex] atoms of Li
- N (Nitrogen):
- Reactants: [tex]\(2\)[/tex] atoms of N
- Products: [tex]\(2\)[/tex] atoms of N
5. Balance Lithium (Li) next:
Since we need 6 lithium atoms on the product side, we should have 6 lithium atoms on the reactant side:
[tex]\[ 6 \text{Li}_{(s)} + \text{N}_2{(g)} \rightarrow 2 \text{Li}_3\text{N}_{(s)} \][/tex]
6. Confirm the balance:
- Li (Lithium):
- Reactants: [tex]\(6\)[/tex] atoms of Li
- Products: [tex]\(3 \times 2 = 6\)[/tex] atoms of Li
- N (Nitrogen):
- Reactants: [tex]\(2\)[/tex] atoms of N
- Products: [tex]\(2\)[/tex] atoms of N
Both lithium and nitrogen are balanced with 6 lithium atoms and 2 nitrogen atoms on both sides of the equation.
Balanced equation:
[tex]\[ 6 \text{Li}_{(s)} + \text{N}_2{(g)} \rightarrow 2 \text{Li}_3\text{N}_{(s)} \][/tex]
Therefore, the coefficients are:
- Lithium ([tex]\(\text{Li}\)[/tex]): 6
- Nitrogen ([tex]\(\text{N}_2\)[/tex]): 1
- Lithium Nitride ([tex]\(\text{Li}_3\text{N}\)[/tex]): 2
This completes the balance of the chemical equation.
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