Under certain conditions, 16g A and 22g B react completely! A chemistry problem. Urgently

According to the first data, it can be seen that the generated C is 16G+ because it is a complete reaction, according to the theorem of conservation of mass) and then the second reaction, B material is excessive, 8GA and 11GB can be completely reacted, and then when 8GA is completely reacted, the mass of C generated is half of the first data, that is, and the condition tells you that the generated C, that is, the molar mass is the mass divided by the amount of matter, and the molar mass is 134, So the relative molecular mass of c is 134.

Since a:b=16:22 is completely reacted, 8 grams of a corresponds to 11 grams of b, yes, 15 grams of b is an excess, which is great, then when 8 grams of a is calculated, and there is a conservation of mass, 16 grams a corresponds to c = 16g+

Then the same proportional, 8 grams of a corresponds to gram c, and the inscription says that this is, then it will come out after dividing it. m(c)=134g/mol

The simplest consideration, according to Equation 2, B is excessive, so the reverse push out of 16Ga+22GB obtains, if AB is just after the reaction, according to the conservation of mass, subtract the mass of D, then generate, the molecular weight of C.

If you want to be more complicated, it's to consider whether the ab overdose reaction is complete or something, I won't go into detail about this, because it will make you dizzy.

Because lead 16a and 20GB happen to react completely, so when 8ga and 15GB react, A reacts completely, according to the law of conservation of mass, the mass-sensitive amount of 16GA and 20GB reaction to produce C is, and the amount of matter is twice that of 15GB reaction to produce C) The relative molecular mass of C is numerically equal to the molar mass of C, so the relative fractional mass of C is.

Mass of Participating Reaction A = 16g-12g = 4g; The amount of quality ridge Kai of b = 16g; The mass of c = 27g-16g = 11g (because the mass increases after the reaction to the wild swim, so it is a product); The mass of d generated at the same time = (4g + 16g) - 11g = 9g;

Multiply each item by 4, a = 16g; b=64g;c=44g;d=36g；So the reaction equation is a+2b=c+2d.

a + 2b = c + 2d

4g 16g 11g 9g 】

The mass is conserved, d is 8+32-22== 18 g16 ga, and the maximum silver can do and 32*2==64 g b.

Now the training is in the 70b then b excessive.

Then the mass d is 18*16 8==36 g, then the molar mass of d == mass The amount of the mass substance is matched with the amount of the substance ===36 2==18 g mol

Under the application of the law of conservation of mass and dispersion, that is, in a chemical reaction, the sum of the masses of the reactants = the sum of the masses of the products Note that the reactants refer to the ones involved in the reaction, and in some topics, it may involve the problem of excess of some reactants, and it is necessary to pay attention to judgment.

In this problem, 8ga and 16GB happen to be the complete reaction of the slow digging (note this condition), then the sum of the masses of the reactants = 8 + 16 = 24 g, and the product has c and d and the mass of c is known to be = 10 g, so the mass of d = a + b - c = 14 g

Then the amount of 12ga participating in the reaction a is a multiple of the previous process, so the mass of d generated in proportion should also be a multiple of the mass of d in the previous disturbance process, i.e. .

a + b = c + d

Cha Hong remembers the good reaction 16 20 (find).

The reaction given is 8 15

The actual reaction circle is void 8 10

From the law of mass constancy, we can know that the resulting c is the gram.

m=m/n=

(1) Assuming that there are 12Ga and 8GB before the reaction, 16GC is generated after the reaction, A is completely consumed, and B has 2G left, then D 2 G is generated;

2) The sum of the masses of the substances participating in the chemical reaction in (1) is 18 g, and the sum of the masses of the substances formed after the reaction is 18 g, which is called the law of conservation of mass

3) (1) In order for B to fully react, a 4 g must be added.

One: 2gTwo: 18g, 18g, the law of conservation of mass. Three: 4g

a + b --c + d

4g 16g 11g(9g)

i.e. a+2b=c+2d

The mass is divided by the relative molecular bench and the amount of the jujube that is measured.

Answer B analysis: According to the law of conservation of mass, in a chemical reaction, the sum of the masses of the substances before participating in the reaction is equal to the sum of the masses of the substances formed after the reaction In this chemical reaction, the mass of A and B is equal to the mass of C and D generated, and the mass of A and B and the mass of C can be obtained from the mass of A and B, so as to know the mass ratio of A and D in this reaction, and then the mass of D can be obtained by the mass of D

Answer: Generate the quality of d:

Let the mass of the substance participating in the reaction a be x

a+2 b═c+d

xfrac=\frac

x = A: The mass of the substance that participates in the reaction is .

Comments: This question mainly tests the ability of students to use the law of conservation of mass to make inferences, and the key to doing the question is to know that the mass ratio of each substance leakage in the same chemical reaction is a fixed value"