_{2}N + 3 O_{2} = 2 CO_{2} + 2 H_{2}O + 2 N_{2}
Direct liên kết lớn this balanced equation: Please tell about this miễn phí chemistry software lớn your friends! | ||||||||||||||||||||||||||||||||||||||||||||

## 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 lớn 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 tự 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 lớn 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 lớn balance:- Fe + Cl
_{2}= FeCl_{3} - KMnO
_{4}+ HCl = KCl + MnCl_{2}+ H_{2}O + Cl_{2} - K
_{4}Fe(CN)_{6}+ H_{2}SO_{4}+ H_{2}O = K_{2}SO_{4}+ FeSO_{4}+ (NH_{4})_{2}SO_{4}+ CO - C
_{6}H_{5}COOH + O_{2}= CO_{2}+ H_{2}O - K
_{4}Fe(CN)_{6}+ KMnO_{4}+ H_{2}SO_{4}= KHSO_{4}+ Fe_{2}(SO_{4})_{3}+ MnSO_{4}+ HNO_{3}+ CO_{2}+ H_{2}O - Cr
_{2}O_{7}{-2} + H{+} + {-} = Cr{+3} + H_{2}O - S{-2} + I
_{2}= I{-} + S - PhCH
_{3}+ KMnO_{4}+ H_{2}SO_{4}= PhCOOH + K_{2}SO_{4}+ MnSO_{4}+ H_{2}O - CuSO
_{4}*5H_{2}O = CuSO_{4}+ H_{2}O - calcium hydroxide + carbon dioxide = calcium carbonate + water
- sulfur + ozone = sulfur dioxide
## Examples of the chemical equations reagents (a complete equation will be suggested):- H
_{2}SO_{4}+ K_{4}Fe(CN)_{6}+ KMnO_{4} - Ca(OH)
_{2}+ H_{3}PO_{4} - Na
_{2}S_{2}O_{3}+ I_{2} - C
_{8}H_{18}+ O_{2} - hydrogen + oxygen
- propane + oxygen
## Understanding chemical equationsA 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) lớn khuông water (H₂O), the chemical equation is: H 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 methodThis is the most straightforward method. It involves looking at the equation and adjusting the coefficients lớn 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:H_{2} + O_{2} = H_{2}O- 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.
- Balance the oxygen atoms by placing a coefficient of 2 in front of H
_{2}O:H _{2}+ O_{2}= 2H_{2}O - Now, there are 4 H atoms on the right side, ví we adjust the left side lớn match:
2H _{2}+ O_{2}= 2H_{2}O - Check the balance. Now, both sides have 4 H atoms and 2 O atoms. The equation is balanced.
## Balancing with algebraic methodThis method uses algebraic equations lớn find the correct coefficients. Each molecule's coefficient is represented by a variable (like x, nó, z), and a series of equations are mix 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 lớn each coefficient, write equations for each element, and then solve the system of equations lớn find the values of the variables. Xem thêm: dấu hiệu nhận biết tứ giác nội tiếp Example: C_{2}H_{6} + O_{2} = CO_{2} + H_{2}O- Assign variables lớn coefficients:
a C _{2}H_{6}+ b O_{2}= c CO_{2}+ d H_{2}O - Write down equations based on atom conservation:
- 2 a = c
- 6 a = 2 d
- 2 b = 2c + d
- Assign one of the coefficients lớn 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
- Adjust coefficient lớn make sure all of them are integers. b = 3.5 ví we need lớn multiple all coefficient by 2 lớn arrive at the balanced equation with integer coefficients:
2 C _{2}H_{6}+ 7 O 2 = 4 CO_{2}+ 6 H_{2}O
## Balancing with oxidation number methodUseful 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 = Ca_{3}P_{2}- Assign oxidation numbers:
- Calcium (Ca) has an oxidation number of 0 in its elemental khuông.
- Phosphorus (P) also has an oxidation number of 0 in its elemental khuông.
- In Ca
_{3}P_{2}, calcium has an oxidation number of +2, and phosphorus has an oxidation number of -3.
- Identify the changes in oxidation numbers:
- Calcium goes from 0 lớn +2, losing 2 electrons (reduction).
- Phosphorus goes from 0 lớn -3, gaining 3 electrons (oxidation).
- Balance the changes using electrons: Multiply the number of calcium atoms by 3 and the number of phosphorus atoms by 2.
- Write the balanced Equation:
3 Ca + 2 Phường = Ca _{3}P_{2}
## Balancing with ion-electron half-reaction methodThis 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 + HNO_{3} = Cu(NO_{3})_{2} + NO_{2} + H_{2}O- Write down and balance half reactions:
Cu = Cu{2+} + 2{e} H{+} + HNO _{3}+ {e} = NO_{2}+ H_{2}O - Combine half reactions lớn balance electrons. To accomplish that we multiple the second half reaction by 2 and add it lớn the first one:
Cu + 2H{+} + 2HNO _{3}+ 2{e} = Cu{2+} + 2NO_{2}+ 2H_{2}O + 2{e} - Cancel out electrons on both sides and add NO
_{3}{-} ions. H{+} with NO_{3}{-} makes HNO_{3}and Cu{2+} with NO_{3}{-} makes Cu(NO_{3})_{3}:Cu + 4HNO _{3}= Cu(NO_{3})_{2}+ 2NO_{2}+ 2H_{2}O
## Learn lớn balance chemical equations:
Xem thêm: bài văn tả cô giáo hay nhất
## Related chemical tools:- Molar mass calculator
- pH solver
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