# fecl3 ra fe(oh)3

Balanced equation:

FeCl3 + 3 NaOH = Fe(OH)3 + 3 NaCl
Reaction type: double replacement
Reaction stoichiometryLimiting reagent
CompoundCoefficientMolar MassMolesWeight
FeCl31162.20
NaOH340.00
Fe(OH)31106.87
NaCl358.44
Units: molar mass - g/mol, weight - g.
Word equation
Iron(III) chloride + 3 Sodium hydroxide = Iron(III) hydroxide + 3 Sodium chloride

Full ionic equation
Fe{+3} + 3 Cl{-} + 3 Na{+} + 3 OH{-} = Fe{+3} + 3 OH{-} + 3 Na{+} + 3 Cl{-}

Balancing step by step using the inspection method
Let's balance this equation using the inspection method.
First, we phối all coefficients to tướng 1:
1 FeCl3 + 1 NaOH = 1 Fe(OH)3 + 1 NaCl

For each element, we kiểm tra if the number of atoms is balanced on both sides of the equation.
Fe is balanced: 1 atom in reagents and 1 atom in products.
Cl is not balanced: 3 atoms in reagents and 1 atom in products.
In order to tướng balance Cl on both sides we:
Multiply coefficient for NaCl by 3
1 FeCl3 + 1 NaOH = 1 Fe(OH)3 + 3 NaCl

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Na is not balanced: 1 atom in reagents and 3 atoms in products.
In order to tướng balance Na on both sides we:
Multiply coefficient for NaOH by 3
1 FeCl3 + 3 NaOH = 1 Fe(OH)3 + 3 NaCl

O is balanced: 3 atoms in reagents and 3 atoms in products.
H is balanced: 3 atoms in reagents and 3 atoms in products.
All atoms are now balanced and the whole equation is fully balanced:
FeCl3 + 3 NaOH = Fe(OH)3 + 3 NaCl

Balancing step by step using the algebraic method
Let's balance this equation using the algebraic method.
First, we phối all coefficients to tướng variables a, b, c, d, ...
a FeCl3 + b NaOH = c Fe(OH)3 + d NaCl

Now we write down algebraic equations to tướng balance of each atom:
Fe: a * 1 = c * 1
Cl: a * 3 = d * 1
Na: b * 1 = d * 1
O: b * 1 = c * 3
H: b * 1 = c * 3

Now we assign a=1 and solve the system of linear algebra equations:
a = c
a * 3 = d
b = d
b = c * 3
b = c * 3
a = 1

Solving this linear algebra system we arrive at:
a = 1
b = 3
c = 1
d = 3

To get to tướng integer coefficients we multiply all variable by 1
a = 1
b = 3
c = 1
d = 3

Now we substitute the variables in the original equations with the values obtained by solving the linear algebra system and arrive at the fully balanced equation:
FeCl3 + 3 NaOH = Fe(OH)3 + 3 NaCl

Direct links to tướng this balanced equation:

### Instructions on balancing chemical equations:

• Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below
• Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide
• To enter an electron into a chemical equation use {-} or e
• To enter an ion, specify charge after the compound in curly brackets: {+3} or {3+} or {3}.
Example: Fe{3+} + I{-} = Fe{2+} + I2
• Substitute immutable groups in chemical compounds to tướng avoid ambiguity.
For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,
but PhC2H5 + O2 = PhOH + CO2 + H2O will
• Compound states [like (s) (aq) or (g)] are not required.
• If you vì thế not know what products are, enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.
• Reaction stoichiometry could be computed for a balanced equation. Enter either the number of moles or weight for one of the compounds to tướng compute the rest.
• Limiting reagent can be computed for a balanced equation by entering the number of moles or weight for all reagents. The limiting reagent row will be highlighted in pink.

### Examples of complete chemical equations to tướng balance:

• Fe + Cl2 = FeCl3
• KMnO4 + HCl = KCl + MnCl2 + H2O + Cl2
• K4Fe(CN)6 + H2SO4 + H2O = K2SO4 + FeSO4 + (NH4)2SO4 + CO
• C6H5COOH + O2 = CO2 + H2O
• K4Fe(CN)6 + KMnO4 + H2SO4 = KHSO4 + Fe2(SO4)3 + MnSO4 + HNO3 + CO2 + H2O
• Cr2O7{-2} + H{+} + {-} = Cr{+3} + H2O
• S{-2} + I2 = I{-} + S
• PhCH3 + KMnO4 + H2SO4 = PhCOOH + K2SO4 + MnSO4 + H2O
• CuSO4*5H2O = CuSO4 + H2O
• calcium hydroxide + carbon dioxide = calcium carbonate + water
• sulfur + ozone = sulfur dioxide

### Examples of the chemical equations reagents (a complete equation will be suggested):

• H2SO4 + K4Fe(CN)6 + KMnO4
• Ca(OH)2 + H3PO4
• Na2S2O3 + I2
• C8H18 + O2
• hydrogen + oxygen
• propane + oxygen

### Understanding chemical equations

A chemical equation represents a chemical reaction. It shows the reactants (substances that start a reaction) and products (substances formed by the reaction). For example, in the reaction of hydrogen (H₂) with oxygen (O₂) to tướng size water (H₂O), the chemical equation is:

H2 + O2 = H2O

However, this equation isn't balanced because the number of atoms for each element is not the same on both sides of the equation. A balanced equation obeys the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction.

### Balancing with inspection or trial and error method

This is the most straightforward method. It involves looking at the equation and adjusting the coefficients to tướng get the same number of each type of atom on both sides of the equation.

Best for: Simple equations with a small number of atoms.

Process: Start with the most complex molecule or the one with the most elements, and adjust the coefficients of the reactants and products until the equation is balanced.

Example:H2 + O2 = H2O
1. Count the number of H and O atoms on both sides. There are 2 H atoms on the left and 2 H atom on the right. There are 2 O atoms on the left and 1 O atom on the right.
2. Balance the oxygen atoms by placing a coefficient of 2 in front of H2O:

H2 + O2 = 2H2O

3. Now, there are 4 H atoms on the right side, so sánh we adjust the left side to tướng match:

2H2 + O2 = 2H2O

4. Check the balance. Now, both sides have 4 H atoms and 2 O atoms. The equation is balanced.

### Balancing with algebraic method

This method uses algebraic equations to tướng find the correct coefficients. Each molecule's coefficient is represented by a variable (like x, nó, z), and a series of equations are phối up based on the number of each type of atom.

Best for: Equations that are more complex and not easily balanced by inspection.

Process: Assign variables to tướng each coefficient, write equations for each element, and then solve the system of equations to tướng find the values of the variables.

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Example: C2H6 + O2 = CO2 + H2O
1. Assign variables to tướng coefficients:

a C2H6 + b O2 = c CO2 + d H2O

2. Write down equations based on atom conservation:
• 2 a = c
• 6 a = 2 d
• 2 b = 2c + d
3. Assign one of the coefficients to tướng 1 and solve the system.
• a = 1
• c = 2 a = 2
• d = 6 a / 2 = 4
• b = (2 c + d) / 2 = (2 * 2 + 3) / 2 = 3.5
4. Adjust coefficient to tướng make sure all of them are integers. b = 3.5 so sánh we need to tướng multiple all coefficient by 2 to tướng arrive at the balanced equation with integer coefficients:

2 C2H6 + 7 O 2 = 4 CO2 + 6 H2O

### Balancing with oxidation number method

Useful for redox reactions, this method involves balancing the equation based on the change in oxidation numbers.

Best For: Redox reactions where electron transfer occurs.

Process: identify the oxidation numbers, determine the changes in oxidation state, balance the atoms that change their oxidation state, and then balance the remaining atoms and charges.

Example: Ca + Phường = Ca3P2
1. Assign oxidation numbers:
• Calcium (Ca) has an oxidation number of 0 in its elemental size.
• Phosphorus (P) also has an oxidation number of 0 in its elemental size.
• In Ca3P2, calcium has an oxidation number of +2, and phosphorus has an oxidation number of -3.
2. Identify the changes in oxidation numbers:
• Calcium goes from 0 to tướng +2, losing 2 electrons (reduction).
• Phosphorus goes from 0 to tướng -3, gaining 3 electrons (oxidation).
3. Balance the changes using electrons: Multiply the number of calcium atoms by 3 and the number of phosphorus atoms by 2.
4. Write the balanced Equation:

3 Ca + 2 Phường = Ca3P2

### Balancing with ion-electron half-reaction method

This method separates the reaction into two half-reactions – one for oxidation and one for reduction. Each half-reaction is balanced separately and then combined.

Best for: complex redox reactions, especially in acidic or basic solutions.

Process: split the reaction into two half-reactions, balance the atoms and charges in each half-reaction, and then combine the half-reactions, ensuring that electrons are balanced.

Example: Cu + HNO3 = Cu(NO3)2 + NO2 + H2O
1. Write down and balance half reactions:

Cu = Cu{2+} + 2{e}

H{+} + HNO3 + {e} = NO2 + H2O

2. Combine half reactions to tướng balance electrons. To accomplish that we multiple the second half reaction by 2 and add it to tướng the first one:

Cu + 2H{+} + 2HNO3 + 2{e} = Cu{2+} + 2NO2 + 2H2O + 2{e}

3. Cancel out electrons on both sides and add NO3{-} ions. H{+} with NO3{-} makes HNO3 and Cu{2+} with NO3{-} makes Cu(NO3)3:

Cu + 4HNO3 = Cu(NO3)2 + 2NO2 + 2H2O

### Related chemical tools:

• Molar mass calculator
• pH solver

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