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A chemical equation is symbolic representation of a chemical reaction where the reactant entities are given on the left hand side and the product entities on the right hand side. The coefficients next to the symbols and formulae of entities are the absolute values of the stoichiometric numbers. The first chemical equation was diagrammed by Jean Beguin in 1615.
A chemical equation consists of the chemical formulas of the reactants (the starting substances) and the chemical formula of the products (substances formed in the chemical reaction). The two are separated by an arrow symbol (\rightarrow, usually read as "yields") and each individual substance's chemical formula is separated from others by a plus sign.
As an example, the formula for the burning of methane can be denoted:
- CH|4| + 2 O|2| \rightarrow CO|2| + 2 H|2|O
This equation would be read as "CH four plus O two produces CO two and H two O." But for equations involving complex chemicals, rather than reading the letter and its subscript, the chemical formulas are read using IUPAC nomenclature. Using IUPAC nomenclature, this equation would be read as "methane plus oxygen yields carbon dioxide and water."
This equation indicates that oxygen and CH4 react to form H2O and CO2. It also indicates that two oxygen molecules are required for every methane molecule and the reaction will form two water molecules and one carbon dioxide molecule for every methane and two oxygen molecules that react. The stoichiometric coefficients (the numbers in front of the chemical formulas) result from the law of conservation of mass and the law of conservation of charge (see "Balancing Chemical Equation" section below for more information).
Symbols are used to differentiate between different types of reactions. To denote the type of reaction:
- "=" symbol is used to denote a stoichiometric relation.
- "\rightarrow" symbol is used to denote a net forward reaction.
- "\rightleftarrows" symbol is used to denote a reaction in both directions.
- "\rightleftharpoons" symbol is used to denote an equilibrium.
Physical state of chemicals is also very commonly stated in parentheses after the chemical symbol, especially for ionic reactions. When stating physical state, (s) denotes a solid, (l) denotes a liquid, (g) denotes a gas and (aq) denotes an aqueous solution.
If the reaction requires energy, it is indicated above the arrow. A capital Greek letter delta (\Delta) is put on the reaction arrow to show that energy in the form of heat is added to the reaction. h\nu is used if the energy is added in the form of light.
Balancing chemical equations
The law of conservation of mass dictates the quantity of each element does not change in a chemical reaction. Thus, each side of the chemical equation must represent the same quantity of any particular element. Similarly, the charge is conserved in a chemical reaction. Therefore, the same charge must be present on both sides of the balanced equation.
One balances a chemical equation by changing the scalar number for each chemical formula. Simple chemical equations can be balanced by inspection, that is, by trial and error. Another technique involves solving a system of linear equations.
Ordinarily, balanced equations are written with smallest whole-number coefficients. If there is no coefficient before a chemical formula, the coefficient 1 is understood.
The method of inspection can be outlined as putting a coefficient of 1 in front of the most complex chemical formula and putting the other coefficients before everything else such that both sides of the arrows have the same number of each atom. If any fractional coefficient exist, multiply every coefficient with the smallest number required to make them whole, typically the denominator of the fractional coefficient for a reaction with a single fractional coefficient.
As an example, the burning of methane would be balanced by putting a coefficient of 1 before the CH4:
- 1 CH|4| + O|2| \rightarrow CO|2| + H|2|O
Since there is one carbon on each side of the arrow, the first atom (carbon) is balanced.
Looking at the next atom (hydrogen), the right hand side has two atoms, while the left hand side has four. To balance the hydrogens, 2 goes in front of the H2O, which yields:
- 1 CH|4| + O|2| \rightarrow CO|2| + 2 H|2|O
Inspection of the last atom to be balanced (oxygen) shows that the right hand side has four atoms, while the left hand side has two. It can be balanced by putting a 2 before O2, giving the balanced equation:
- CH|4| + 2 O|2| \rightarrow CO|2| + 2 H|2|O
This equation does not have any coefficients in front of CH4 and CO2, since a coefficient of 1 is dropped.
An ionic equation is a chemical equation in which electrolytes are written as dissociated ions. Ionic equations are used for single and double displacement reactions that occur in aqueoussolutions. For example in the following precipitation reaction:
- CaCl2(aq) + 2AgNO3(aq) \rightarrow Ca(NO3)2(aq) + 2AgCl(s)
the full ionic equation would be:
- Ca2+ + 2Cl− + 2Ag+ + 2NO3− \rightarrow Ca2+ + 2NO3− + 2AgCl(s)
and the net ionic equation would be:
- 2Cl−(aq) + 2Ag+(aq) \rightarrow 2AgCl(s)
or, in reduced balanced form,
- Ag+ + Cl− \rightarrow AgCl(s)
In this aqueous reaction the Ca2+ an
From Yahoo Answers
Answers:Balancing chemical equations is a difficult skill to master. I will try to explain it as well as I can: Let's start with aerobic respiration: C6H12O6 + O2 --> H2O +CO2 I am going to name the coefficients to make it easier to follow. (A) C6H12O6 + (B) O2 --> (C) H2O + (D) CO2 The first thing I always look for is an element that is in just one reactant and one product. In this case, Hydrogen comes only from glucose and ends up only in water. Because we have 12 hydrogen atoms in glucose and 2 in water, we know there must be 6 times as many water molecules as glucose molecules. Thus, 6C=A We can also see that Carbon atoms come only from glucose and end up only in carbon dioxide. Because there are 6 times as many atoms in one molecule of glucose as there are in one molecule of carbon dioxide, we also know that 6D=A Since 6C=A=6D we know that 6C=6D which means C=D We still haven't balanced oxygen, but let's just try putting in 1 for C and D and 6 for A. 1 C6H12O6 + (B) O2 --> 6 H2O + 6 CO2 We are trying to find B. On the left side of the equation, 6 atoms of oxygen come from glucose. On the right side of the equation, 6 atoms of oxygen are in water and 12 are in carbon dioxide. Thus, we know that the left side needs a total of 12 more oxygen atoms to be balanced. Thus, there must be 6 Oxygen molecules to balance the equation. Our final balanced equation is then: C6H12O6 + 6 O2 --> 6 H2O + 6 CO2 To check our answer we just total up the atoms on each side. ----------------------------------------------------------------- This was the thought process that I would go through to balance this equation. I hope this helps a bit. If you ever get stuck, just try guessing a number and see if it works. Here is another example that is more difficult: CH2S + F2 CF4 + HF + SF6 Again I will name the coefficients: (A) CH2S + (B) F2 (C) CF4 + (D) HF + (E) SF6 Let's assume there is one molecule of CH2S on the left side: 1 CH2S + (B) F2 (C) CF4 + (D) HF + (E) SF6 Looking at the right side, we see that the only compound that contains carbon is CF4. Thus, there must also be 1 molecule of CF4: 1 CH2S + (B) F2 1 CF4 + (D) HF + (E) SF6 Like Carbon, there is only one compound that contains hydrogen on the right side, HF. Because there is 1 atom of Hydrogen in HF and 2 in CH2S, we know there must be twice as much HF as CH2S. 1 CH2S + (B) F2 1 CF4 + 2 HF + (E) SF6 Again, we have one atom of Sulfur on the left, so there must be one also on the right. The only compound that contains sulfur on the right is SF6. We must therefore have one molecule of SF6 to balance the sulfur. 1 CH2S + (B) F2 1 CF4 + 2 HF + 1 SF6 Now the only thing left to balance is Fluorine. There are zero fluorine atoms in the compound CH2S. Looking at the right side, there are 4 in CF4, 1 in each molecule of HF and 6 in SF6. Thus we have a total of 4+2*1+6=12. Thus, we must add 12 fluorine to the left side. Because there are two atoms of fluorine in each molecule of F2, we must add 6 molecules of F2 to the left side. Our final balanced equation is: 1 CH2S + 6 F2 1 CF4 + 2 HF + 1 SF6 I hope this helps. If you ever get stuck, just guess a number for a single compound and then balance the others based on that.
Answers:I'm doing this too first of all write your equation out which would be ___, then write the product. You should know that. it's very easy to balance, after you do that you can work on your stoicheomtry
Answers:3) CaO(s) + H2O(l) ==> Ca(OH)2(aq) This is a combination reaction. 4) CO2(g) + H2O(l) ==> H2CO3(aq) ...another combination reaction. 5) NiO(s) + 2HNO3(aq) ==> H2O(l) + Ni(NO3)2(aq) This is a double replacement reaction.
Answers:1. Alumin + Hydochloric Acid 2Al + 6HCl = 2AlCl3 + 3H2 [Single displacement] 2. Calcium hydroxide + nitric acid Ca(OH)2 + 2HNO3 = Ca(NO3)2 + 2H2O [Double displacement] 3. Potassium chlorate heated 2KClO3 = 2KCl + 3O2 [decomposition] 4. Magnesium + sulfur (S8) 4Mg + S = Mg4S [synthesis] 5. Ammonium phosphate + alumin Chloride (NH4)3PO4 + AlCl3 = AlPO4 + 3NH4Cl [double replacement] 6. Calcium oxide + water CaO + H2O = Ca(OH)2 [Synthesis] 7. Chromium + water Cr + 6H2O = Cr(H2O)6 [synthesis] 8. Tin + mercury(I) nitrate Sn + 2HgNO3 = Sn(NO3)2 + 2Hg [Single displacement] 9. Sodium Bromide + silver nitrate NaBr + AgNO3 = NaNO3 + AgBr [double replacement] 10. hydrogen + oxygen 2H2 + O2 = 2H2O [Synthesis] 11. hydogen peroxide H2O2 = H2 + O2 [decomposition] 12. fluorine + potasium bromide F + KBr = FBr + K [single displacement] 13. Carbon dioxide + water CO2 + H2O = H2CO3 [Synthesis] 14. calcium chloride + ammonium hydroxide CaCl2 + 2NH4OH = Ca(OH)2 + 2NH4Cl [double replacement] 15. Sodium + chlorine 2Na + Cl2 = 2NaCl [synthesis] 16. bromine + sodium chloride Br2 + 2NaCl = 2NaBr + Cl2 [single replacement] 17. mercury (II) oxide heated 2HgO = 2Hg + O2 [decomposition] 18. potassium + water 2K + 2H2O = 2KOH + H2 [SINGLE replacement] 19. strontium carbonate + nitric acid SrCO3 + 2HNO3 = Sr(NO3)2 + H2CO3 [double replacement] 20. potassium iodide + lead nitrate 2KI + Pb(NO3)2 = 2KNO3 + PbI2 [double replacement] :) whoo! that's a lot.