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Balance the following chemical equations AICI3 + H2O = AL(OH)3 + HCI

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To balance the chemical equation \( \text{AlCl}_3 + \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + \text{HCl} \), follow these steps:

  1. Write down the unbalanced equation: \[ \text{AlCl}_3 + \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + \text{HCl} \]

  2. Count the number of each type of atom on both sides:

    • Reactants:
      • Al: 1 (from AlCl3)
      • Cl: 3 (from AlCl3)
      • H: 2 (from H2O)
      • O: 1 (from H2O)
    • Products:
      • Al: 1 (from Al(OH)3)
      • Cl: 1 (from HCl)
      • H: 4 (3 from Al(OH)3 + 1 from HCl)
      • O: 3 (from Al(OH)3)

  3. Start balancing the equation:

    • Balance the number of chlorine atoms by placing a coefficient of 3 in front of HCl: \[ \text{AlCl}_3 + \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + 3 \text{HCl} \]

  4. Now recount and balance the hydrogen and oxygen:

    • On the product side, we now have:
      • H: 3 (from Al(OH)3) + 3 (from 3 HCl) = 6
      • O: 3 (from Al(OH)3)

  5. To balance the hydrogen, place a coefficient of 3 in front of water: \[ \text{AlCl}_3 + 3 \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + 3 \text{HCl} \]

  6. Now recount the atoms:

    • Reactants:
      • Al: 1
      • Cl: 3
      • H: 6 (from 3 H2O)
      • O: 3 (from 3 H2O)
    • Products:
      • Al: 1
      • Cl: 3
      • H: 6 (3 from Al(OH)3 and 3 from 3 HCl)
      • O: 3 (from Al(OH)3)

Now everything is balanced. The final balanced equation is:

\[ \text{AlCl}_3 + 3 \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + 3 \text{HCl} \]

C3H8 + O2 = CO2 + H2O

To balance the combustion reaction of propane (\( \text{C}_3\text{H}_8 \)), we start with the unbalanced equation:

\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]

Steps to Balance the Equation:

  1. Count the Atoms:

    • Reactants:
      • C: 3 (from \( \text{C}_3\text{H}_8 \))
      • H: 8 (from \( \text{C}_3\text{H}_8 \))
      • O: 2 (from \( \text{O}_2 \))
    • Products:
      • C: 1 (from \( \text{CO}_2 \))
      • H: 2 (from \( \text{H}_2\text{O} \))
      • O: 3 (2 from \( \text{CO}_2 \) and 1 from \( \text{H}_2\text{O} \))

  2. Balance Carbon Atoms: Place a coefficient of 3 in front of \( \text{CO}_2 \): \[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + \text{H}_2\text{O} \]

  3. New Count of Products:

    • Products:
      • C: 3 (from 3 \( \text{CO}_2 \))
      • H: 2 (from \( \text{H}_2\text{O} \))
      • O: 7 (3 from 3 \( \text{CO}_2 \) and 1 from \( \text{H}_2\text{O} \))

  4. Balance Hydrogen Atoms: Place a coefficient of 4 in front of \( \text{H}_2\text{O} \): \[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]

  5. Recount the Atoms:

    • Products:
      • C: 3
      • H: 8 (4 from \( \text{H}_2\text{O} \))
      • O: 10 (6 from 3 \( \text{CO}_2 \) and 4 from 4 \( \text{H}_2\text{O} \))

  6. Balance Oxygen Atoms: We now have 10 oxygen atoms in the products. Since \( O_2 \) has 2 oxygen atoms per molecule, we need 5 \( O_2 \) to provide 10 oxygen atoms: \[ \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]

Final Balanced Equation:

\[ \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]

This is the balanced equation for the combustion of propane.

AL + FeO = AI2O3 + Fe

To balance the chemical equation \( \text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe} \), follow these steps:

  1. Write the unbalanced equation: \[ \text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe} \]

  2. Count the number of each type of atom on both sides:

    • Reactants:
      • Al: 1
      • Fe: 1
      • O: 1
    • Products:
      • Al: 2 (from \( \text{Al}_2\text{O}_3 \))
      • Fe: 1
      • O: 3 (from \( \text{Al}_2\text{O}_3 \))

  3. Balance aluminum atoms: To balance the aluminum atoms, we need 2 Al in the reactants, so we place a coefficient of 2 in front of Al: \[ 2 \text{Al} + \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe} \]

  4. Recount the atoms:

    • Reactants:
      • Al: 2
      • Fe: 1
      • O: 1
    • Products:
      • Al: 2
      • Fe: 1
      • O: 3

  5. Balance oxygen atoms: To balance the oxygen atoms, we know we need 3 oxygen atoms on the reactant side. Since each \( \text{FeO} \) provides 1 oxygen atom, we need 3 \( \text{FeO} \): \[ 2 \text{Al} + 3 \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + \text{Fe} \]

  6. Balance iron atoms: Now, with 3 \( \text{FeO} \), we have 3 iron atoms on the reactant side. Therefore, we should have a coefficient of 3 in front of \( \text{Fe} \) in the products: \[ 2 \text{Al} + 3 \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3 \text{Fe} \]

  7. Final balanced equation: After ensuring all atoms are balanced, we get: \[ 2 \text{Al} + 3 \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3 \text{Fe} \]

Summary:

The balanced equation is: \[ 2 \text{Al} + 3 \text{FeO} \rightarrow \text{Al}_2\text{O}_3 + 3 \text{Fe} \]

S8 + O2 = SO3

To balance the combustion reaction of sulfur (\( \text{S}_8 \)) with oxygen (\( \text{O}_2 \)) to form sulfur trioxide (\( \text{SO}_3 \)), follow these steps:

  1. Write the unbalanced equation: \[ \text{S}_8 + \text{O}_2 \rightarrow \text{SO}_3 \]

  2. Count the number of each type of atom on both sides:

    • Reactants:
      • S: 8 (from \( \text{S}_8 \))
      • O: 2 (in \( \text{O}_2 \))
    • Products:
      • S: 1 (from \( \text{SO}_3 \))
      • O: 3 (from \( \text{SO}_3 \))

  3. Balance sulfur atoms: To balance the sulfur atoms, we need 8 sulfur atoms in the products. Therefore, place a coefficient of 8 in front of \( \text{SO}_3 \): \[ \text{S}_8 + \text{O}_2 \rightarrow 8 \text{SO}_3 \]

  4. Recount the atoms:

    • Products now:
      • S: 8 (from \( 8 \text{SO}_3 \))
      • O: 24 (from \( 8 \text{SO}_3 \); since \( 8 \times 3 = 24 \))

  5. Balance oxygen atoms: We now have 24 oxygen atoms in the products and \( \text{O}_2 \) in the reactants, which means we need 12 \( \text{O}_2 \) molecules because each \( \text{O}_2 \) contains 2 oxygen atoms: \[ \text{S}_8 + 12 \text{O}_2 \rightarrow 8 \text{SO}_3 \]

  6. Final balanced equation: After confirming all atoms balance, we have: \[ \text{S}_8 + 12 \text{O}_2 \rightarrow 8 \text{SO}_3 \]

Summary:

The balanced equation is: \[ \text{S}_8 + 12 \text{O}_2 \rightarrow 8 \text{SO}_3 \]

C7H6O2 + O2 = CO2 + H2O

To balance the combustion reaction of benzoic acid (\( \text{C}_7\text{H}_6\text{O}_2 \)) with oxygen (\( \text{O}_2 \)), we start with the unbalanced equation:

\[ \text{C}_7\text{H}_6\text{O}_2 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]

Steps to Balance the Equation:

  1. Count the Atoms in the Reactants:

    • Carbon (C): 7 (from \( \text{C}_7\text{H}_6\text{O}_2 \))
    • Hydrogen (H): 6 (from \( \text{C}_7\text{H}_6\text{O}_2 \))
    • Oxygen (O): 2 (from \( \text{C}_7\text{H}_6\text{O}_2 \)) + \( x \) (from \( \text{O}_2 \))

  2. Balance Carbon Atoms: Place a coefficient of 7 in front of \( \text{CO}_2 \): \[ \text{C}_7\text{H}_6\text{O}_2 + \text{O}_2 \rightarrow 7 \text{CO}_2 + \text{H}_2\text{O} \]

  3. Count the Atoms in the Products:

    • From \( 7 \text{CO}_2 \):
      • Carbon (C): 7
      • Oxygen (O): \( 7 \times 2 = 14 \) (from \( \text{CO}_2 \))
    • Hydrogen (H): Needs to be balanced, so place a coefficient of 3 in front of \( \text{H}_2\text{O} \): \[ \text{C}_7\text{H}_6\text{O}_2 + \text{O}_2 \rightarrow 7 \text{CO}_2 + 3 \text{H}_2\text{O} \]

  4. Count the Atoms in the Products Again:

    • Total in Products:
      • C: 7 (from \( 7 \text{CO}_2 \))
      • H: 6 (from \( 3 \text{H}_2\text{O} \) since \( 3 \times 2 = 6 \))
      • O: \( 14 + 3 = 17 \) (from \( 7 \text{CO}_2 \) and \( 3 \text{H}_2\text{O} \))

  5. Calculate the Needed Oxygen: The total number of oxygen atoms required on the reactant side is 17. The reactant \( \text{C}_7\text{H}_6\text{O}_2 \) already has 2 oxygen atoms, so we need \( 17 - 2 = 15 \) more oxygen atoms from \( \text{O}_2 \): \[ 15 \text{ O} / 2 = 7.5 \text{ O}_2 \]

  6. Modify for Whole Numbers: It's preferable to use whole numbers for coefficients. To eliminate the fraction, multiply the entire equation by 2: \[ 2 \text{C}_7\text{H}_6\text{O}_2 + 15 \text{O}_2 \rightarrow 14 \text{CO}_2 + 6 \text{H}_2\text{O} \]

Final Balanced Equation:

The final balanced equation for the combustion of benzoic acid is: \[ 2 \text{C}_7\text{H}_6\text{O}_2 + 15 \text{O}_2 \rightarrow 14 \text{CO}_2 + 6 \text{H}_2\text{O} \]

BaF2 + K3PO4 = Ba3(PO4)2 +KF

To balance the reaction between barium fluoride (\( \text{BaF}_2 \)) and potassium phosphate (\( \text{K}_3\text{PO}_4 \)) to form barium phosphate (\( \text{Ba}_3(\text{PO}_4)_2 \)) and potassium fluoride (\( \text{KF} \)), follow these steps:

  1. Write the unbalanced equation: \[ \text{BaF}_2 + \text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + \text{KF} \]

  2. Count the number of each type of atom on both sides:

    • Reactants:

      • Ba: 1 (from \( \text{BaF}_2 \))
      • F: 2 (from \( \text{BaF}_2 \))
      • K: 3 (from \( \text{K}_3\text{PO}_4 \))
      • P: 1 (from \( \text{K}_3\text{PO}_4 \))
      • O: 4 (from \( \text{K}_3\text{PO}_4 \))

    • Products:

      • Ba: 3 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
      • F: 1 (from \( \text{KF} \))
      • K: 1 (from \( \text{KF} \))
      • P: 2 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
      • O: 8 (from \( \text{Ba}_3(\text{PO}_4)_2 \))

  3. Balance barium (Ba) first: To balance barium, put a coefficient of 3 in front of \( \text{BaF}_2 \): \[ 3\text{BaF}_2 + \text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + \text{KF} \]

  4. Recount the atoms:

    • Reactants:

      • Ba: 3 (from \( 3\text{BaF}_2 \))
      • F: 6 (from \( 3\text{BaF}_2 \))
      • K: 3 (from \( \text{K}_3\text{PO}_4 \))
      • P: 1 (from \( \text{K}_3\text{PO}_4 \))
      • O: 4 (from \( \text{K}_3\text{PO}_4 \))

    • Products:

      • Ba: 3 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
      • F: 1 (from \( \text{KF} \))
      • K: 1 (from \( \text{KF} \))
      • P: 2 (from \( \text{Ba}_3(\text{PO}_4)_2 \))
      • O: 8 (from \( \text{Ba}_3(\text{PO}_4)_2 \))

  5. Balance phosphate (P): To account for the 2 phosphate groups in the product, we need to have 2 \( \text{K}_3\text{PO}_4 \): \[ 3\text{BaF}_2 + 2\text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + \text{KF} \]

  6. Balance water: As we balance potassium from \( \text{K}_3\text{PO}_4 \) (which contributes 6 K total), we need to balance that with \( \text{KF} \): \[ 3\text{BaF}_2 + 2\text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + 6 \text{KF} \]

Final balanced equation:

After all adjustments, the fully balanced equation is: \[ 3\text{BaF}_2 + 2\text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + 6\text{KF} \]

Summary:

The balanced reaction maintains the same number of atoms for each element on both sides of the equation.