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Balance the following chemical equations

FeCI3 + NaOH = Fe(OH)3 + NaCI

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To balance the chemical equation for the reaction between iron(III) chloride (FeCl₃) and sodium hydroxide (NaOH) to form iron(III) hydroxide (Fe(OH)₃) and sodium chloride (NaCl), follow these steps:

  1. Write the unbalanced equation: \[ \text{FeCl}_3 + \text{NaOH} \rightarrow \text{Fe(OH)}_3 + \text{NaCl} \]

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

    • Reactants:

      • Fe: 1 in FeCl₃
      • Cl: 3 in FeCl₃
      • Na: 1 in NaOH
      • O: 1 in NaOH
      • H: 1 in NaOH

    • Products:

      • Fe: 1 in Fe(OH)₃
      • Cl: 1 in NaCl
      • Na: 1 in NaCl
      • O: 3 in Fe(OH)₃
      • H: 3 in Fe(OH)₃

  3. Balance the equation by adjusting the coefficients:

    • Iron (Fe) is balanced with 1 on each side.
    • Chlorine (Cl) has 3 on the left (in FeCl₃) and only 1 on the right (in NaCl), so we need 3 NaCl: \[ \text{FeCl}_3 + \text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaCl} \]
    • Now we have 3 Na in the products, so put a coefficient of 3 for NaOH: \[ \text{FeCl}_3 + 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaCl} \]

  4. Now recount the atoms:

    • Reactants:
      • Fe: 1
      • Cl: 3
      • Na: 3
      • O: 3
      • H: 3
    • Products:
      • Fe: 1
      • Cl: 3
      • Na: 3
      • O: 3
      • H: 3

Both sides are balanced.

The balanced equation is: \[ \text{FeCl}_3 + 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaCl} \]

K3PO4 + HCI = KCI + H3PO4

To balance the chemical equation for the reaction between potassium phosphate (K₃PO₄) and hydrochloric acid (HCl) to form potassium chloride (KCl) and phosphoric acid (H₃PO₄), follow these steps:

  1. Write the unbalanced equation: \[ \text{K}_3\text{PO}_4 + \text{HCl} \rightarrow \text{KCl} + \text{H}_3\text{PO}_4 \]

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

    • Reactants:

      • K: 3 (in K₃PO₄)
      • P: 1 (in K₃PO₄)
      • O: 4 (in K₃PO₄)
      • H: 1 (in HCl)
      • Cl: 1 (in HCl)

    • Products:

      • K: 1 (in KCl)
      • Cl: 1 (in KCl)
      • H: 3 (in H₃PO₄)
      • P: 1 (in H₃PO₄)
      • O: 4 (in H₃PO₄)

  3. Balance the equation:

    • Potassium (K): There are 3 K atoms in the reactants, so we need to put a coefficient of 3 in front of KCl: \[ \text{K}_3\text{PO}_4 + \text{HCl} \rightarrow 3\text{KCl} + \text{H}_3\text{PO}_4 \]

    • Now recount the reactants:

      • K: 3 (in K₃PO₄)
      • H: 1 (in HCl)
      • Cl: 3 (from 3 KCl)

    • To balance the 3 HCl needed, we now adjust the coefficient in front of HCl to 3: \[ \text{K}_3\text{PO}_4 + 3\text{HCl} \rightarrow 3\text{KCl} + \text{H}_3\text{PO}_4 \]

  4. Final count:

    • Reactants:

      • K: 3
      • P: 1
      • O: 4
      • H: 3
      • Cl: 3

    • Products:

      • K: 3
      • Cl: 3
      • H: 3
      • P: 1
      • O: 4

Both sides are balanced.

The final balanced equation is: \[ \text{K}_3\text{PO}_4 + 3\text{HCl} \rightarrow 3\text{KCl} + \text{H}_3\text{PO}_4 \]

CaBr2 + Na3P = Ca3P2 + NaBr

To balance the chemical equation for the reaction between calcium bromide (CaBr₂) and sodium phosphate (Na₃P) to form calcium phosphate (Ca₃P₂) and sodium bromide (NaBr), follow these steps:

  1. Write the unbalanced equation: \[ \text{CaBr}_2 + \text{Na}_3\text{P} \rightarrow \text{Ca}_3\text{P}_2 + \text{NaBr} \]

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

    • Reactants:

      • Ca: 1 (in CaBr₂)
      • Br: 2 (in CaBr₂)
      • Na: 3 (in Na₃P)
      • P: 1 (in Na₃P)

    • Products:

      • Ca: 3 (in Ca₃P₂)
      • P: 2 (in Ca₃P₂)
      • Br: 1 (in NaBr)
      • Na: 1 (in NaBr)

  3. Balance the equation:

    • Start by balancing calcium (Ca):

      • Since there are 3 Ca in the products (Ca₃P₂), we need 3 CaBr₂ in the reactants: \[ 3\text{CaBr}_2 + \text{Na}_3\text{P} \rightarrow \text{Ca}_3\text{P}_2 + \text{NaBr} \]

    • Now, recount:

      • Reactants:

        • Ca: 3
        • Br: 6 (from 3 CaBr₂)
        • Na: 3 (from Na₃P)
        • P: 1

      • Products:

        • Ca: 3
        • P: 2
        • Br: 1 (from NaBr)
        • Na: 1

    • Next, balance phosphorus (P). Since we have 2 P in the products (Ca₃P₂), we need to have 2 Na₃P: \[ 3\text{CaBr}_2 + 2\text{Na}_3\text{P} \rightarrow \text{Ca}_3\text{P}_2 + \text{NaBr} \]

      • Now recount:

      • Reactants:

        • Ca: 3
        • Br: 6 (3 CaBr₂)
        • Na: 6 (2 Na₃P)
        • P: 2 (2 Na₃P)

      • Products:

        • Ca: 3
        • P: 2
        • Br: 1 (from NaBr)
        • Na: 1 (from NaBr)

    • Finally, balance sodium (Na):

      • Since we have 6 Na in the reactants, we need to have 6 NaBr in the products: \[ 3\text{CaBr}_2 + 2\text{Na}_3\text{P} \rightarrow \text{Ca}_3\text{P}_2 + 6\text{NaBr} \]

  4. Final count:

    • Reactants:

      • Ca: 3
      • Br: 6
      • Na: 6 (from 2 Na₃P)
      • P: 2 (from 2 Na₃P)

    • Products:

      • Ca: 3 (from Ca₃P₂)
      • P: 2 (from Ca₃P₂)
      • Na: 6 (from 6 NaBr)
      • Br: 6 (from 6 NaBr)

Both sides are balanced.

The final balanced equation is: \[ 3\text{CaBr}_2 + 2\text{Na}_3\text{P} \rightarrow \text{Ca}_3\text{P}_2 + 6\text{NaBr} \]

B2Br6 + HNO3 = B(NO3)e + HBr

To balance the chemical equation for the reaction between dibromine (B₂Br₆) and nitric acid (HNO₃) to form boron nitrate (B(NO₃)₃) and hydrobromic acid (HBr), we can follow these steps:

  1. Write the unbalanced equation: \[ \text{B}_2\text{Br}_6 + \text{HNO}_3 \rightarrow \text{B(NO}_3)_3 + \text{HBr} \]

  2. Identify and count the number of atoms on both sides:

    • Reactants:

      • B: 2 (in B₂Br₆)
      • Br: 6 (in B₂Br₆)
      • H: 1 (in HNO₃)
      • N: 1 (in HNO₃)
      • O: 3 (in HNO₃)

    • Products:

      • B: 1 (in B(NO₃)₃)
      • Br: 1 (in HBr)
      • H: 1 (in HBr)
      • N: 3 (in B(NO₃)₃, one for each nitrate group)
      • O: 9 (in B(NO₃)₃, three nitrate groups each containing three oxygen atoms)

  3. Balance the equation:

    • Start with boron (B):
      • Since we have 2 B in the reactants, we need 2 B(NO₃)₃ in the products: \[ \text{B}_2\text{Br}_6 + \text{HNO}_3 \rightarrow 2\text{B(NO}_3)_3 + \text{HBr} \]
    • Now recount:
      • Products now have:
        • B: 2 (from 2 B(NO₃)₃)
        • N: 6 (from 2 B(NO₃)₃)
        • H: 1 (from HBr)
        • O: 18 (from 2 B(NO₃)₃)
        • Br: 1 (from HBr)
      • In order to balance the nitrogen and hydrobromic acid, we need to put in the appropriate coefficients.

  4. Determine how many nitric acid and hydrobromic acid are needed:

  • There are 6 nitrogen atoms in total from 2 B(NO₃)₃. Thus, we need 6 HNO₃: \[ \text{B}_2\text{Br}_6 + 6\text{HNO}_3 \rightarrow 2\text{B(NO}_3)_3 + \text{HBr} \]

  • For each bromine atom in B₂Br₆, we need a corresponding HBr, which means we need 6 HBr to balance the 6 Br: \[ \text{B}_2\text{Br}_6 + 6\text{HNO}_3 \rightarrow 2\text{B(NO}_3)_3 + 6\text{HBr} \]

  1. Final count:

    • Reactants:

      • B: 2 (from B₂Br₆)
      • Br: 6 (from B₂Br₆)
      • H: 6 (from 6 HNO₃)
      • N: 6 (from 6 HNO₃)
      • O: 18 (from 6 HNO₃)

    • Products:

      • B: 2 (from 2 B(NO₃)₃)
      • Br: 6 (from 6 HBr)
      • H: 6 (from 6 HBr)
      • N: 6 (from 2 B(NO₃)₃)
      • O: 18 (from 2 B(NO₃)₃)

Now both sides are balanced.

The final balanced equation is: \[ \text{B}_2\text{Br}_6 + 6\text{HNO}_3 \rightarrow 2\text{B(NO}_3)_3 + 6\text{HBr} \]