Chemistry Questions Detailed Steps Chemistry Questions To detailed steps

1. From the equation CaCO3+2HCl==CaCl2+CO2+H2O, it can be seen that grams of carbon dioxide are collected, and the mass of calcium carbonate participating in the reaction is 40 grams, so the mass fraction of limestone is 80%;

2. From the equation H2SO4+2NaOH==Na2SO4+2H2O, it can be seen that the mass of sulfuric acid participating in the reaction is 49 grams, 40 grams of sodium hydroxide need to be consumed, and if the mass fraction of sodium hydroxide in the waste liquid is 2%, 2 kg of waste liquid is required;

3. Calculated by the mass fraction of nitrogen (for example, the mass fraction of nitrogen in urea is 14*2 (12+16+28+4)=, it can be seen that the nitrogen content of urea is the highest, and the nitrogen content in 10 kg of urea is about kilograms.

CaCO3+2HCl=CaCl2+H2O+CO2x column proportional formula: 100 x=44 Solution, the mass fraction of x=40g calcium carbonate is: 40 50*100%=80%49g x column proportional formula:

80 x = 98 49 solution, get x = 40g40 2% = 2000g = 2kg

3.Urea has the highest nitrogen content; Ammonium bicarbonate has the lowest nitrogen content; (28/60)*10=

I'm thinking about that too.

The third sub-question is what it means.

There are clo2 and clo2- in the water, first use i- to convert clo2 to clo2- titration of the generated i2 to know the amount of clo2, then i2 changes back to i- to adjust ph, and then use i- to convert clo2- to cl-

The titration of the generated I2 knows that the difference between the total amount of Clo2- (generated plus the original) is the amount of the original Clo2-.

That's the idea.

I don't understand why the ratio of Na2SO3 Clo2 to Clo2 is 1:1 when converting Clo2 to Clo2-

Answer: (1) The ClO2 prepared in method 2 does not contain Cl2 (2) ClO2 4H4i = Cl 2I2 2H2O to adjust the pH of the solution

C(V2-4V1) 4V Fe(OH)3 Analysis: (1) It is relatively simple, observe the two sets of equations, and see the product combined with its properties to get the answer. (2) At pH, clo2 can also be completely reduced to cl by i, then i should be oxidized to iodine element, and the solution is an acidic solution, and the ion equation can be written correctly.

The information prompt of the topic is to use the continuous iodine measurement method for determination, and the step 6 is titrated to the end point with Na2S2O3 solution, indicating that there is iodine generation in step 5, combined with When please write out the pH, the ion equation of the reaction between Clo2 and i and the presence of Clo in the solution with pH can be used to determine the pH of the adjustment solution. To find the concentration of Clo2 in the cited water sample, the relationship between Clo2 and Na2S2O3 can be correctly identified by the relational method, using iodine as a bridge. Fe2+ reduces Clo2 to Cl, Fe2 is oxidized to iron ions, and it is not difficult to find the answer by combining reactants.

Sodium, magnesium, aluminum.

The analysis and solution of such questions should not be done in conventional ways, but should start from the valency and relative archetypal confessional quality. The product of the mass of the metal and the valency is removed relative to the mass of the vertical atom, and the resulting value is the mass of the hydrogen produced.

If you think about it yourself, you will be more impressed (you are lazy and not verbose).

c can find that the condition of 1 is the slowest.

So d corresponds to 1

Note 3,4

It is impossible to judge under the existing conditions, only on the basis of 1,2.

C is chosen because 1 has the smallest concentration, the lowest temperature, and is still lumpy, so its reaction rate is the smallest, that is, the maximum volume value is reached at the latest.

（1）ch4+2h2oco2+4h2

3h2+n2 2nh3

2nh3+co2===co（nh2）2+h2och4：

4 000 mol

ch4～4h2～8/3nh3

The amount of substances used to synthesize ammonia is: 4 000 mol 8 3 60% = 6 400 mol

2nh3～co（nh2）2

The quality of the production of urea is:

6 400 mol×1/2×60 g·mol-1=192 000 g=192 kg

2）4nh3+5o2 4no+6h2o

2no+o2===2no2

3no2+2h2o===2hno3+no

The nitrogen in ammonia is all converted to Hno3, then 5NH3 10O2V(NH3) V(air)=5 10 20%=1 10(3)NH3+HNO3===NH4NO3

Let the amount of ammonia used in the production of nitric acid be x

The mass of 6 400 mol-x=x·80% x=3 molNH4NO3 is 3 mol 80% 80 g·mol-1=227 g= kg

<> can you see it?

I'm writing a bit messy.

Four nitrogen atoms

An N4 molecule

Two nitrogen molecules

- Energy level diagram ---

The N4 molecule, which has six single bonds, breaks all the bonds to convert into four nitrogen atoms to absorb 193x6 = 1158 kJ mol

N4 --4N δH = 1158 kJ mol Two N2 molecules, two triplets, break all the bonds and convert into four nitrogen atoms to absorb 941x2 = 1882kJ mol

2N2 --4N δH = 1882kJ mol from the simple energy diagram above, on....

N4 --4N δH=1158kJ mol) 4n --2N2 ΔH=-1882kJ molN4 --2N2 ΔH=-724kJ mol, so it is exothermic.

N4 contains 4 N-N bonds, and the energy required: 193*4=772kJ mol

The N2 molecule is formed, which emits an energy of 941kj mol so N4 is converted to N2 and will be exothermic.

There are 2 scenarios.

1. When the amount of NaOH is greater than the amount of NaHCO3 substance, the reaction is:

NaOH + NaHCO3 == Na2CO3 + H2O product only H2O, for, so NaHCO3 is, NaOH is 10g, and the mass fraction is 10

2. When the amount of NaOH substance is less than the amount of NaHCO3 substance, the reaction is:

NaOH + NaHCO3 == Na2CO3 + H2O2NaHCO3 == Na2CO3 + H2O + CO2 The products are CO2 and H2O

Let the quantities of NaOH and NaHCO3 be x mol and y mol respectively, respectively, to obtain the system of equations: 40x + 84y =

18x + 62（y-x）/2 =

The result is x>y

If you don't make assumptions, you should discard it.

The reduced substance is Ho + Co =

The total molecular weight of HO+Co is 18+44=622NaHCO==== NaCo+HO+CoX

x = mass fraction of NaOH in the original mixture = (

Solution: Heat.

2NaHCO3***** Na2CO3+H2O+CO2, the remaining solid is a mixture of NaOH and Na2CO3, and the gas is H2O and CO2, then the mass of H2O and CO2 is.

Let NaHCO3 have a mass of m.

It is calculated that m=

The key to solving chemistry problems is to understand the meaning:

1. "Put it in a closed container and heat it to 300", first of all, you have to know what reaction will occur under this condition, sodium hydroxide will not react, and 2nahco == == na co + h o + co.

2. What is the gas discharged from the "exhaust gas after sufficient reaction"? Since it is heated to 300 meters, both water and carbon dioxide are gases and are discharged. Why is there only residue after cooling?

It's because water and carbon dioxide are discharged. What is reduced is the total mass of water and carbon dioxide produced. Now know that the total mass of water and carbon dioxide produced is.

What is the mass of water or carbon dioxide can be found according to the ratio of water to carbon dioxide, and then directly substituted into the chemical equation to solve. I'm not going to demonstrate it here. Other buildings have already given the correct answer.

(1)、v（b）=

2) The amount of substance d according to the title is.

So the amount of substances that react a and b is and is respectively.

Let the amount of substance starting a be x mol

x = 3 starts with the amount of a substance of 3mol

(1) VB: VC = 1:2 (rate ratio = coefficient ratio) so VB = , 2) Let the amount of substance A at the beginning be X The concentration of D is , and the closed container of 2L generates 1mol, then ABC reacts respectively: ; 21

3a(g)+b(g)===2cg)+2d(g), starting x x

Reaction +1 +1 (:(5 (the ratio of concentrations given, because the volume is the same, it can also be the ratio of the amount of the substance) gives x=3mol