QUESTION IMAGE
Question
- ____ naoh → __ na₂o + ____ h₂o
| reactants | products | |
|---|---|---|
| o | ||
| h |
- ____ al₂o₃ → __ al + ____ o₂
| reactants | products | |
|---|---|---|
| o |
- ____ k₂co₃ → __ k₂o + ____ co₂
| reactants | products | |
|---|---|---|
| c | ||
| o |
- ____ zn(oh)₂ → __ zno + ____ h₂o
| reactants | products | |
|---|---|---|
| o | ||
| h |
- ____ ag₂o → __ ag + ____ o₂
| reactants | products | |
|---|---|---|
| o |
- ____ ni(clo₃)₂ → __ nicl₂ + ____ o₂
| reactants | products | |
|---|---|---|
| cl | ||
| o |
- ____ hcl → __ h₂ + ____ cl₂
| reactants | products | |
|---|---|---|
| cl |
Step 1: Balance Equation 7
- Reactants: NaOH has 1 Na, 1 O, 1 H.
- Products: \( Na_2O \) has 2 Na, 1 O; \( H_2O \) has 2 H, 1 O.
- To balance Na: Put 2 in front of NaOH. Now NaOH: 2 Na, 2 O, 2 H.
- Check O and H: \( Na_2O \) (1 O) + \( H_2O \) (1 O) = 2 O; \( H_2O \) has 2 H. So coefficients: 2 NaOH → 1 \( Na_2O \) + 1 \( H_2O \).
- Reactants (Na: 2, O: 2, H: 2); Products (Na: 2, O: 2, H: 2).
Step 2: Balance Equation 8
- Reactants: \( Al_2O_3 \) has 2 Al, 3 O.
- Products: Al (1 Al), \( O_2 \) (2 O).
- Balance Al: Put 2 in front of Al. Now Al: 2.
- Balance O: \( Al_2O_3 \) has 3 O, \( O_2 \) needs 3 O (but \( O_2 \) is diatomic). Multiply \( Al_2O_3 \) by 2 (4 Al, 6 O) and Al by 4, \( O_2 \) by 3. So 2 \( Al_2O_3 \) → 4 Al + 3 \( O_2 \).
- Reactants (Al: 4, O: 6); Products (Al: 4, O: 6).
Step 3: Balance Equation 9
- Reactants: \( K_2CO_3 \) has 2 K, 1 C, 3 O.
- Products: \( K_2O \) (2 K, 1 O), \( CO_2 \) (1 C, 2 O).
- K: 2 on both. C: 1 on both. O: 3 (1 + 2) on products. So coefficients: 1 \( K_2CO_3 \) → 1 \( K_2O \) + 1 \( CO_2 \).
- Reactants (K: 2, C: 1, O: 3); Products (K: 2, C: 1, O: 3).
Step 4: Balance Equation 10
- Reactants: \( Zn(OH)_2 \) has 1 Zn, 2 O, 2 H.
- Products: ZnO (1 Zn, 1 O), \( H_2O \) (2 H, 1 O).
- Zn: 1 on both. O: 2 (1 + 1) on products. H: 2 on both. So coefficients: 1 \( Zn(OH)_2 \) → 1 ZnO + 1 \( H_2O \).
- Reactants (Zn: 1, O: 2, H: 2); Products (Zn: 1, O: 2, H: 2).
Step 5: Balance Equation 11
- Reactants: \( Ag_2O \) has 2 Ag, 1 O.
- Products: Ag (1 Ag), \( O_2 \) (2 O).
- Balance Ag: Put 2 in front of Ag. Now Ag: 2.
- Balance O: \( Ag_2O \) has 1 O, \( O_2 \) needs 2 O. Multiply \( Ag_2O \) by 2 (4 Ag, 2 O) and Ag by 4, \( O_2 \) by 1. So 2 \( Ag_2O \) → 4 Ag + 1 \( O_2 \).
- Reactants (Ag: 4, O: 2); Products (Ag: 4, O: 2).
Step 6: Balance Equation 12
- Reactants: \( Ni(ClO_3)_2 \) has 1 Ni, 2 Cl, 6 O.
- Products: \( NiCl_2 \) (1 Ni, 2 Cl), \( O_2 \) (2 O).
- Ni and Cl: balanced (1 Ni, 2 Cl).
- Balance O: 6 O in reactants. \( O_2 \) needs 3 (6/2). So coefficients: 1 \( Ni(ClO_3)_2 \) → 1 \( NiCl_2 \) + 3 \( O_2 \).
- Reactants (Ni: 1, Cl: 2, O: 6); Products (Ni: 1, Cl: 2, O: 6).
Step 7: Balance Equation 13
- Reactants: HCl has 1 H, 1 Cl.
- Products: \( H_2 \) (2 H), \( Cl_2 \) (2 Cl).
- Balance H: Put 2 in front of HCl. Now HCl: 2 H, 2 Cl.
- Products: \( H_2 \) (1) and \( Cl_2 \) (1). So coefficients: 2 HCl → 1 \( H_2 \) + 1 \( Cl_2 \).
- Reactants (H: 2, Cl: 2); Products (H: 2, Cl: 2).
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- 2 NaOH → 1 \( Na_2O \) + 1 \( H_2O \)
Reactants: Na=2, O=2, H=2; Products: Na=2, O=2, H=2
- 2 \( Al_2O_3 \) → 4 Al + 3 \( O_2 \)
Reactants: Al=4, O=6; Products: Al=4, O=6
- 1 \( K_2CO_3 \) → 1 \( K_2O \) + 1 \( CO_2 \)
Reactants: K=2, C=1, O=3; Products: K=2, C=1, O=3
- 1 \( Zn(OH)_2 \) → 1 ZnO + 1 \( H_2O \)
Reactants: Zn=1, O=2, H=2; Products: Zn=1, O=2, H=2
- 2 \( Ag_2O \) → 4 Ag + 1 \( O_2 \)
Reactants: Ag=4, O=2; Products: Ag=4, O=2
- 1 \( Ni(ClO_3)_2 \) → 1 \( NiCl_2 \) + 3 \( O_2 \)
Reactants: Ni=1, Cl=2, O=6; Products: Ni=1, Cl=2, O=6
- 2 HCl → 1 \( H_2 \) + 1 \( Cl_2 \)
Reactants: H=2, Cl=2; Products: H=2, Cl=2