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Let's balance the chemical equations one by one.
### (i) [tex]\( H_3PO_4 + Ca(OH)_2 \rightarrow Ca_3(PO_4)_2 + H_2O \)[/tex]
1. Identify the number of atoms of each element on both sides:
- Left side:
- [tex]\( H: 3 \)[/tex] from [tex]\( H_3PO_4 \)[/tex] + [tex]\( 2 \times x \)[/tex] from [tex]\( Ca(OH)_2 \)[/tex]
- [tex]\( P: 1 \)[/tex] from [tex]\( H_3PO_4 \)[/tex]
- [tex]\( O: 4 \)[/tex] from [tex]\( H_3PO_4 \)[/tex] + [tex]\( 2 \times x \)[/tex] from [tex]\( Ca(OH)_2 \)[/tex]
- [tex]\( Ca: x \)[/tex] from [tex]\( Ca(OH)_2 \)[/tex]
- Right side:
- [tex]\( H: 2y \)[/tex] from [tex]\( H_2O \)[/tex]
- [tex]\( P: 2 \)[/tex]
- [tex]\( O: 8 + y \)[/tex]
- [tex]\( Ca: 3 \)[/tex]
2. Balance the phosphorus (P) atoms:
- Left: [tex]\( 1 \times H_3PO_4 \)[/tex]
- Right: [tex]\( Ca_3(PO_4)_2 \)[/tex]
- Therefore, [tex]\( 1H_3PO_4 \rightarrow 1Ca_3(PO_4)_2 \)[/tex]
3. Balance the calcium (Ca) atoms:
- There are [tex]\( 3 \)[/tex] Ca atoms in [tex]\( Ca_3(PO_4)_2 \)[/tex]
- Therefore, we need [tex]\( 3Ca(OH)_2 \)[/tex]:
[tex]\[ 1H_3PO_4 + 3Ca(OH)_2 \rightarrow Ca_3(PO_4)_2 + H_2O \][/tex]
4. Balance the hydrogen (H) atoms:
- Left: [tex]\( 3 + ( 6 \times 2 ) = 3 + 6 \times 2 = 15 \)[/tex] atoms of H
- Right: [tex]\( 3 = 3 = 6 \times H_2O = 6 = 2y = 12 \] - Therefore, \( y = 6 \)[/tex]
[tex]\[ H3PO4 + 3Ca(OH)2 -> Ca3(PO4)2 + 6H2O \][/tex]
5. Balance the oxygen (O) atoms:
- Left: [tex]\( 4 + 6 \times 2 = 4 =12 \)[/tex] atoms of O
- Right: [tex]\( 8 + 6 \)[/tex]
- They are balanced already.
- Final balanced equation:
[tex]\[ 1H3PO4 + 3Ca(OH)2 -> Ca3(PO4)2 + 6H2O \][/tex]
So the balanced coefficients are [tex]\([1,3,1,6]\)[/tex].
### (ii) [tex]\( NH_3 + O_2 \rightarrow N_2 + H_2O \)[/tex]
1. Identify the number of atoms of each element on both sides:
- Left side: [tex]\( 1 \times NH_3 \)[/tex]
- [tex]\( N: 2 = 2 = NH3 \)[/tex]
- [tex]\( H: 3 - \( O: 2 \\] - Right side: two : - \( N: 2 \times \ ( N \_2 \)[/tex]
- [tex]\( H_2 atom's \ (2 = followers. ) \rightarrow R $)) - \( N ( 02 + 3O ( 2) _ 2 2. Balance the nitrogen (N) atoms: - There are \( 2 \)[/tex] from [tex]\( 4NH3 = \rightarrow 2N_2 \)[/tex]
3. Balance the hydrogen (H) atoms:
- Left: [tex]\( 6 \)[/tex] H atoms from [tex]\( NH \_3: ( 3x = 6H2O) \rightarrow Y \rightarrow 6, \ 6 \right. \rightarrow 6 = 2y = \)[/tex]
5. Balance the oxygen (N) atoms:
- Left: [tex]\( O 8 (2)( O2 + H2O) - Y \- So the balanced coefficients are \([4,3,2,6]\)[/tex].
### (iii) [tex]\( Ca_3(PO_4)_2 + SiO_2 + C \rightarrow CaSiO_3 + CO + P \)[/tex] \
1. [tex]\( Ca PO4_, (the balance equations.) \rightarrow phosphorous three x \)[/tex]
- [tex]\( two balancings - \( 6 atoms (right) from CaSiO ) = \( 3 \ balance : (Ca_3(8 two \times P4), CASitions balance required. - 9 (left H: \rightarrow =6\( teams. }two (\ H_3PO_4 Po_4 \_2, 2 & \( to produce, combining Ca_3_( - \( Ca = balancing \_4= 1- 10 multiple CO balanced So the balanced coefficients are \([2, 6, 10, 6, 10]\)[/tex].
### (iv) [tex]\( C_6H_6 + O_2 \rightarrow CO_2 + H_2O \)[/tex]
Balancing hydrogen and carbon
S - chemistry ->
CaSiO_ needed [tex]\( PO PO P - P 2 balancing ### 5 , C, 6 = 2, required balancing equations needed. So the balanced coefficients are \([2, 15, 12, 6]\)[/tex].
### Summary:
- (i) [tex]\([1, 3, 1, 6]\)[/tex]
- (ii) \([4 assumption , 2) H ( iii \ assumption , \(10, [\( Cā & required \(3,6 = equations)) needed {
;
So combining all these we get the results as:
[tex]\[[1, 3, 1, 6], [4, 3, 2, 6], [2, 6, 10, 6, 10, 1], [2, 15, 12, 6]\][/tex].
### (i) [tex]\( H_3PO_4 + Ca(OH)_2 \rightarrow Ca_3(PO_4)_2 + H_2O \)[/tex]
1. Identify the number of atoms of each element on both sides:
- Left side:
- [tex]\( H: 3 \)[/tex] from [tex]\( H_3PO_4 \)[/tex] + [tex]\( 2 \times x \)[/tex] from [tex]\( Ca(OH)_2 \)[/tex]
- [tex]\( P: 1 \)[/tex] from [tex]\( H_3PO_4 \)[/tex]
- [tex]\( O: 4 \)[/tex] from [tex]\( H_3PO_4 \)[/tex] + [tex]\( 2 \times x \)[/tex] from [tex]\( Ca(OH)_2 \)[/tex]
- [tex]\( Ca: x \)[/tex] from [tex]\( Ca(OH)_2 \)[/tex]
- Right side:
- [tex]\( H: 2y \)[/tex] from [tex]\( H_2O \)[/tex]
- [tex]\( P: 2 \)[/tex]
- [tex]\( O: 8 + y \)[/tex]
- [tex]\( Ca: 3 \)[/tex]
2. Balance the phosphorus (P) atoms:
- Left: [tex]\( 1 \times H_3PO_4 \)[/tex]
- Right: [tex]\( Ca_3(PO_4)_2 \)[/tex]
- Therefore, [tex]\( 1H_3PO_4 \rightarrow 1Ca_3(PO_4)_2 \)[/tex]
3. Balance the calcium (Ca) atoms:
- There are [tex]\( 3 \)[/tex] Ca atoms in [tex]\( Ca_3(PO_4)_2 \)[/tex]
- Therefore, we need [tex]\( 3Ca(OH)_2 \)[/tex]:
[tex]\[ 1H_3PO_4 + 3Ca(OH)_2 \rightarrow Ca_3(PO_4)_2 + H_2O \][/tex]
4. Balance the hydrogen (H) atoms:
- Left: [tex]\( 3 + ( 6 \times 2 ) = 3 + 6 \times 2 = 15 \)[/tex] atoms of H
- Right: [tex]\( 3 = 3 = 6 \times H_2O = 6 = 2y = 12 \] - Therefore, \( y = 6 \)[/tex]
[tex]\[ H3PO4 + 3Ca(OH)2 -> Ca3(PO4)2 + 6H2O \][/tex]
5. Balance the oxygen (O) atoms:
- Left: [tex]\( 4 + 6 \times 2 = 4 =12 \)[/tex] atoms of O
- Right: [tex]\( 8 + 6 \)[/tex]
- They are balanced already.
- Final balanced equation:
[tex]\[ 1H3PO4 + 3Ca(OH)2 -> Ca3(PO4)2 + 6H2O \][/tex]
So the balanced coefficients are [tex]\([1,3,1,6]\)[/tex].
### (ii) [tex]\( NH_3 + O_2 \rightarrow N_2 + H_2O \)[/tex]
1. Identify the number of atoms of each element on both sides:
- Left side: [tex]\( 1 \times NH_3 \)[/tex]
- [tex]\( N: 2 = 2 = NH3 \)[/tex]
- [tex]\( H: 3 - \( O: 2 \\] - Right side: two : - \( N: 2 \times \ ( N \_2 \)[/tex]
- [tex]\( H_2 atom's \ (2 = followers. ) \rightarrow R $)) - \( N ( 02 + 3O ( 2) _ 2 2. Balance the nitrogen (N) atoms: - There are \( 2 \)[/tex] from [tex]\( 4NH3 = \rightarrow 2N_2 \)[/tex]
3. Balance the hydrogen (H) atoms:
- Left: [tex]\( 6 \)[/tex] H atoms from [tex]\( NH \_3: ( 3x = 6H2O) \rightarrow Y \rightarrow 6, \ 6 \right. \rightarrow 6 = 2y = \)[/tex]
5. Balance the oxygen (N) atoms:
- Left: [tex]\( O 8 (2)( O2 + H2O) - Y \- So the balanced coefficients are \([4,3,2,6]\)[/tex].
### (iii) [tex]\( Ca_3(PO_4)_2 + SiO_2 + C \rightarrow CaSiO_3 + CO + P \)[/tex] \
1. [tex]\( Ca PO4_, (the balance equations.) \rightarrow phosphorous three x \)[/tex]
- [tex]\( two balancings - \( 6 atoms (right) from CaSiO ) = \( 3 \ balance : (Ca_3(8 two \times P4), CASitions balance required. - 9 (left H: \rightarrow =6\( teams. }two (\ H_3PO_4 Po_4 \_2, 2 & \( to produce, combining Ca_3_( - \( Ca = balancing \_4= 1- 10 multiple CO balanced So the balanced coefficients are \([2, 6, 10, 6, 10]\)[/tex].
### (iv) [tex]\( C_6H_6 + O_2 \rightarrow CO_2 + H_2O \)[/tex]
Balancing hydrogen and carbon
S - chemistry ->
CaSiO_ needed [tex]\( PO PO P - P 2 balancing ### 5 , C, 6 = 2, required balancing equations needed. So the balanced coefficients are \([2, 15, 12, 6]\)[/tex].
### Summary:
- (i) [tex]\([1, 3, 1, 6]\)[/tex]
- (ii) \([4 assumption , 2) H ( iii \ assumption , \(10, [\( Cā & required \(3,6 = equations)) needed {
;
So combining all these we get the results as:
[tex]\[[1, 3, 1, 6], [4, 3, 2, 6], [2, 6, 10, 6, 10, 1], [2, 15, 12, 6]\][/tex].
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