Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
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!
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
