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Sagot :
Sure, let's go through each reaction step-by-step and classify them accordingly:
### a. Balancing and Classification of [tex]$Ba + N_2 \rightarrow Ba_3N_2$[/tex]
Balancing the reaction:
1. Identify the number of atoms of each element on both sides of the equation.
2. Balance the barium (Ba) atoms first since nitrogen (N) is in a diatomic form.
On the left: 1 Ba, 2 N
On the right: 3 Ba, 2 N
3. The reaction can be balanced by ensuring the number of Ba atoms is the same on both sides. We notice that we need 3 Ba atoms on the left:
[tex]$3Ba + N_2 \rightarrow Ba_3N_2$[/tex]
This is now balanced.
Classification:
This reaction combines barium and nitrogen to form a compound, barium nitride. Hence, it is a synthesis (SY) reaction.
### b. Balancing and Classification of [tex]$Ag + H_2S \rightarrow Ag_2S + H_2$[/tex]
Balancing the reaction:
1. Identify the number of atoms of each element on both sides of the equation.
On the left: 1 Ag, 2 H, 1 S
On the right: 2 Ag, 2 H, 1 S
2. We need to have 2 Ag atoms on the left to balance it with the 2 Ag atoms on the right:
[tex]$2Ag + H_2S \rightarrow Ag_2S + H_2$[/tex]
This is now balanced.
Classification:
This reaction involves a single element (Ag) replacing another element (H) in a compound (H_2S). Hence, it is a single replacement (SR) reaction.
### c. Balancing and Classification of [tex]$CaCl_2 + Na_3PO_4 \rightarrow Ca_3(PO_4)_2 + NaCl$[/tex]
Balancing the reaction:
1. Identify the number of atoms of each element on both sides of the equation.
On the left: 1 Ca, 2 Cl; 3 Na, 1 PO_4 (4 O and 1 P)
On the right: 3 Ca, 2 PO_4 (8 O and 2 P); 1 Na, 1 Cl
2. Balance one compound at a time. Start with Ca and PO_4. To balance the PO_4 groups, we need 2 Na_3PO_4 on the left and 3 CaCl_2 on the left:
[tex]$3CaCl_2 + 2Na_3PO_4 \rightarrow Ca_3(PO_4)_2 + 6NaCl$[/tex]
3. Now check all elements:
On the left: 3 Ca, 6 Cl; 6 Na, 2 PO_4 (8 O and 2 P)
On the right: 3 Ca, 6 Cl; 6 Na, 2 PO_4 (8 O and 2 P)
This is now balanced.
Classification:
This reaction exchanges parts between two compounds (CaCl_2 and Na_3PO_4) to form two new compounds (Ca_3(PO_4)_2 and NaCl). Hence, it is a double replacement (DR) reaction.
### d. Balancing and Classification of [tex]$NaClO_3 \rightarrow NaCl + O_2$[/tex]
Balancing the reaction:
1. Identify the number of atoms of each element on both sides of the equation.
On the left: 1 Na, 1 Cl, 3 O
On the right: 1 Na, 1 Cl, (Oxygen exists as O_2, so consider O2 molecules)
2. Balance the oxygen atoms. Notice that O_2 means each molecule contains 2 Oxygen atoms, so adjust the multiplier to account for the total of 3 oxygen atoms:
[tex]$2NaClO_3 \rightarrow 2NaCl + 3O_2$[/tex]
3. Now check all elements:
On the left: 2 Na, 2 Cl, 6 O
On the right: 2 Na, 2 Cl, 6 O
This is now balanced.
Classification:
This reaction breaks down a single compound (NaClO_3) into two different substances (NaCl and O_2). Hence, it is a decomposition (D) reaction.
### Summary
a. [tex]$ 3Ba + N_2 \rightarrow Ba_3N_2 $[/tex] - Synthesis (SY)
b. [tex]$ 2Ag + H_2S \rightarrow Ag_2S + H_2 $[/tex] - Single Replacement (SR)
c. [tex]$ 3CaCl_2 + 2Na_3PO_4 \rightarrow Ca_3(PO_4)_2 + 6NaCl $[/tex] - Double Replacement (DR)
d. [tex]$ 2NaClO_3 \rightarrow 2NaCl + 3O_2 $[/tex] - Decomposition (D)
Feel free to ask if you have any questions or need further clarification!
### a. Balancing and Classification of [tex]$Ba + N_2 \rightarrow Ba_3N_2$[/tex]
Balancing the reaction:
1. Identify the number of atoms of each element on both sides of the equation.
2. Balance the barium (Ba) atoms first since nitrogen (N) is in a diatomic form.
On the left: 1 Ba, 2 N
On the right: 3 Ba, 2 N
3. The reaction can be balanced by ensuring the number of Ba atoms is the same on both sides. We notice that we need 3 Ba atoms on the left:
[tex]$3Ba + N_2 \rightarrow Ba_3N_2$[/tex]
This is now balanced.
Classification:
This reaction combines barium and nitrogen to form a compound, barium nitride. Hence, it is a synthesis (SY) reaction.
### b. Balancing and Classification of [tex]$Ag + H_2S \rightarrow Ag_2S + H_2$[/tex]
Balancing the reaction:
1. Identify the number of atoms of each element on both sides of the equation.
On the left: 1 Ag, 2 H, 1 S
On the right: 2 Ag, 2 H, 1 S
2. We need to have 2 Ag atoms on the left to balance it with the 2 Ag atoms on the right:
[tex]$2Ag + H_2S \rightarrow Ag_2S + H_2$[/tex]
This is now balanced.
Classification:
This reaction involves a single element (Ag) replacing another element (H) in a compound (H_2S). Hence, it is a single replacement (SR) reaction.
### c. Balancing and Classification of [tex]$CaCl_2 + Na_3PO_4 \rightarrow Ca_3(PO_4)_2 + NaCl$[/tex]
Balancing the reaction:
1. Identify the number of atoms of each element on both sides of the equation.
On the left: 1 Ca, 2 Cl; 3 Na, 1 PO_4 (4 O and 1 P)
On the right: 3 Ca, 2 PO_4 (8 O and 2 P); 1 Na, 1 Cl
2. Balance one compound at a time. Start with Ca and PO_4. To balance the PO_4 groups, we need 2 Na_3PO_4 on the left and 3 CaCl_2 on the left:
[tex]$3CaCl_2 + 2Na_3PO_4 \rightarrow Ca_3(PO_4)_2 + 6NaCl$[/tex]
3. Now check all elements:
On the left: 3 Ca, 6 Cl; 6 Na, 2 PO_4 (8 O and 2 P)
On the right: 3 Ca, 6 Cl; 6 Na, 2 PO_4 (8 O and 2 P)
This is now balanced.
Classification:
This reaction exchanges parts between two compounds (CaCl_2 and Na_3PO_4) to form two new compounds (Ca_3(PO_4)_2 and NaCl). Hence, it is a double replacement (DR) reaction.
### d. Balancing and Classification of [tex]$NaClO_3 \rightarrow NaCl + O_2$[/tex]
Balancing the reaction:
1. Identify the number of atoms of each element on both sides of the equation.
On the left: 1 Na, 1 Cl, 3 O
On the right: 1 Na, 1 Cl, (Oxygen exists as O_2, so consider O2 molecules)
2. Balance the oxygen atoms. Notice that O_2 means each molecule contains 2 Oxygen atoms, so adjust the multiplier to account for the total of 3 oxygen atoms:
[tex]$2NaClO_3 \rightarrow 2NaCl + 3O_2$[/tex]
3. Now check all elements:
On the left: 2 Na, 2 Cl, 6 O
On the right: 2 Na, 2 Cl, 6 O
This is now balanced.
Classification:
This reaction breaks down a single compound (NaClO_3) into two different substances (NaCl and O_2). Hence, it is a decomposition (D) reaction.
### Summary
a. [tex]$ 3Ba + N_2 \rightarrow Ba_3N_2 $[/tex] - Synthesis (SY)
b. [tex]$ 2Ag + H_2S \rightarrow Ag_2S + H_2 $[/tex] - Single Replacement (SR)
c. [tex]$ 3CaCl_2 + 2Na_3PO_4 \rightarrow Ca_3(PO_4)_2 + 6NaCl $[/tex] - Double Replacement (DR)
d. [tex]$ 2NaClO_3 \rightarrow 2NaCl + 3O_2 $[/tex] - Decomposition (D)
Feel free to ask if you have any questions or need further clarification!
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