Correlation method is a clever application of calculation based on chemical equations. The core idea of solving the problem is the conservation of mass in chemical reactions, and there is the most basic proportion (quantity) relationship between reactants and products.
Second, the equation or system of equations method
According to the conservation of mass and proportional relationship, it is the most commonly used method in chemical calculation to set unknowns and solve equations or equations according to set conditions, and its problem-solving skills are also the most important calculation skills.
Third, the conservation method
Because chemical equations can express the quantitative relationship between the quantity, mass and gas volume of substances between reactants and products, it is inevitable that the number of atoms, charges, gain and loss electrons and total mass are conserved before and after chemical reaction. Clever application of conservation law can often simplify the steps of solving problems, solve problems accurately and quickly, and get twice the result with half the effort.
Fourth, the difference method
Find out the difference before and after chemical reaction and the essence and relationship of this difference, and list the method to solve the proportional formula, which is the difference method. The difference may be poor quality, air volume, pressure, etc.
The essence of difference method is to skillfully use calculation according to chemical equations. Its greatest advantage is that as long as the difference is found, the amount consumed by each reactant or the amount produced by each product can be found.
Five, the average method
The average method is a clever problem-solving method, and it is also an important problem-solving thinking and method.
Six, extreme value method
The method of applying mathematical limit knowledge skillfully to chemical calculation is extreme value method.
Seven, cross method
If A and B are used to represent the amounts of two components of binary mixture, the total amount of mixture is A+B (for example, mol).
If xa and xb are used to represent the characteristic quantities (such as molecular weight) of the two components respectively, and X represents the characteristic quantity (such as average molecular weight) of the mixture, then:
Cross method is a special method in the calculation of binary mixture (or composition), which evolved from the calculation of binary linear equation. If the two groups of components and the average value of these two quantities are known, the proportional relationship between these two quantities can be calculated by crossing method.
The key to using the cross method is to conform to the binary linear equation relationship. What calculation is it used for?
Clearly using the cross method to calculate the conditions can list binary linear equations, and special attention should be paid to avoid the abuse of unclear chemical meaning.
The crossover method is mainly used for:
① Calculation of atomic number ratio of two isotopes.
② Calculation of mixture composition and average formula.
③ Calculation of mixed hydrocarbon composition. (Senior 2 content)
④ Calculation of component mass fraction or solution dilution, etc.
Example 7 It is known that there are two isotopes of iridium in nature, the mass numbers are 19 1 and 193 respectively, while the average atomic weight of iridium is 192.22, and the atomic number ratio of these two isotopes should be [].
A.39∶6 1
C. 1∶ 1
This problem can be solved by binary linear equations, but the crossover method is the fastest:
Eight, discussion method
Discussion is a way to discover thinking. When solving calculation problems, if the problem setting conditions are sufficient, you can directly calculate and solve them; If the conditions are not sufficient, we should use the methods of discussion, calculation and reasoning to solve the problem.
Example 8 Fill a 30mL measuring cylinder with the mixed gas of NO2 and O2, stand upside down in water to make the gas fully react, and finally leave 5mL of gas. What is the volume of oxygen in the original mixed gas?
The last 5mL gas may be O2 or NO, which needs to be discussed and analyzed.
Solution (1) The last remaining 5mL gas may be O2; It may also be negative. If it is NO, NO2 exceeds the standard 15mL.
Let the volumes of NO2 and O2 in 30ml original gas mixture be x and y, respectively.
4NO2+O2+2H2O=4HNO3
The volume of oxygen in the original mixed gas may be 10mL or 3mL.
Solution (2):
Let the volume of oxygen in the original mixture be y(mL)
(1) Let O2 be excessive: according to 4NO2+O2+2H2O = 4NO3, the number of electrons in O2 is equal to the number of electrons lost by NO2.
(y-5)×4=(30-y)× 1
Y= 10 (ml)
(2) If NO2 is excessive:
4NO2+O2+2H2O=4HNO3
4y y
Nitric acid +H2O= nitric acid+nitric acid
Because in all (30-y)mLNO2, 5ml NO2 is converted into NO, and the rest (30-y-5)ml NO2 is converted into HNO3, with no electrons.
The number of electrons in O2 +(NO2→NO) is equal to the number of electrons lost when (NO2→HNO3).
Evaluation solution (2) According to the conservation of gain and loss electrons, Avon Gadereau's law is used to transform information, and the volume number is converted into the quantity of matter to simplify the calculation. All redox reactions can generally be calculated by the conservation of electron gain and loss.
No matter the solution (1) or the solution (2), it is necessary to combine the discussion method to solve the problem because of insufficient conditions.
4y+5×2=(30-y-5)× 1
Y=3 (ml)
The volume of raw oxygen can be10ml or 3ml.