Chemistry problems are solved in detail, and the detailed process of solving chemistry problems

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.

On the substitution of the formula, the promotion is set up to be smiling late stool Li Zhishu according to the conditions of the question.

I am glad to answer for you, (1), according to the conservation of iron, the iron in the ore has been transferred to the extracted ferric oxide, and it is known that the relative mass of ferric oxide is 160, then the mass of iron in the ore is, then the mass fraction of iron in the ore is. (2) In ores, the oxide of iron is massive. The mass fraction of iron in oxide is:

Then, in ferrous oxide, the mass fraction of iron is 56 (56+16)*100%=, in ferric oxide it is 56*2 (56*2+16*3)*100%=70%, and in ferric oxide it is 56*3 (56*3+16*4)=, so the oxide in the ore is ferric oxide. Hope I can help you.

Solution: (1) Iron mass =

Iron mass fraction =

2) The mass fraction of iron oxide in the ore is, that is, the chemical formula of oxide mass = iron oxide is composed of FeO, Fe2O3 mixture, FeO molecular weight = Fe2O3 molecular weight =

If the FeO content is x(mol) and the Fe2O3 content is y(mol), then x=,y= are obtained

i.e. x y=1

So the molecular formula is: FeO·Fe2O3 or written as Fe3O4

Solution: (1) The mass of iron:, the mass fraction of iron:

2) The amount of iron in the substance: the amount of oxygen in the substance: (, because n(fe): n(o)=3:4, so the chemical formula of the iron oxide in the iron ore is fe3o4

The Fe element in the iron ore is eventually all converted to Fe2O3, and the relative molecular mass of Fe2O3 is 160, so it is.

So the element iron has, for, and the mass fraction is.

The oxide of iron is iron oxide or ferrous oxide, and the iron oxide xmol and ferrous oxide ymol iron element are conserved to obtain, 2x+y=

Conservation of mass yields 160x+72y=5*

Simultaneous equations can be used to solve the answer.

2na + h2o ==2naoh + h2...1)2al + 2naoh + 2h2o ==2naalo2 + 3h2...2)

m(al) = n(al) = =

n2(h2) generated by reaction (2) = *3 = total n(h2) = =

So the n1(h2) = - = generated by the inverse pin should be (1), so the n(na) = * 2 = of Senyou in the mixture, and because there is still n(naoh) = * 2 = so n(na) +2 * n(na2o) = + = so n(na2o) =

So the mass fraction of sodium primitives in this sodium block = 23 * 23 * 2 * 62 * =

g for water containing h for mol; o for.

The weight gain gram of soda lime is CO2 containing c is , and o is cuo mass reduction indicating that o atoms are lost, and the amount of matter is =, so co; o for.

And because of the carbon monoxide, CO2, and water vapor generated, the total O atom is and contains the O element (

Therefore, there is an organic substance containing o for.

So this organic substance contains h atom , c atom, o atom so the chemical formula of this organic substance is c3h6o3

By increasing the weight of concentrated sulfuric acid and gramming --- water, and by increasing the weight of alkali lime by gramming --- CO2, the mass of cuo decreases --o, reacted, co, o

Carbon monoxide, CO2, water vapor contains, indicating that organic matter does not contain O,N (organic matter): n(c):n(h) = :(:3:6 organic matter chemical formula C3H6

Analysis: After the combustion of mol organic matter and mol O2, the amount of H2O generated is g, that is, mol, the generated CO is: N(CO)== mol, the generated CO2 is: mol= mol, and the amount of O atomic matter in mol organic matter is obtained by using the conservation of oxygen element

moln(c)∶n(h)∶n(o)=(

The molecular formula of the organic matter is C3H6O3.

The amount of N2O5 species at equilibrium is.

So the amount of N2O5 species reflected is, and the amount of N2O3 species formed is so the amount of O2 species formed by the first reaction is.

Because the amount of total O2 substance is 5mol

So the amount of O2 species formed by the second reaction is.

So the second reaction, the amount of N2O3 species reacted is.

So the amount of N2O3 species remaining is.

Because the container is 2L, the N2O3 concentration is:

We can use the method of conservation of elements, which makes the calculation simpler.

Originally, there were a total of 4mol N2O5, and the equilibrium had reaction consumption that is, N, 13mol of O equilibrium to produce O2, 5mol of O, that is, 10mol of O, and 3mol of O in the products N2O3 and N2O.

When equilibrium is set there is n2o3 amol n2o bmol, so n is conserved 2a+2b=

o Conservation 3a+b=3

The solution is a= b=

That is, when the equilibrium N2O3 is there, then the concentration is.

The equilibrium concentration of N2O5 is, then the remaining N2O5 is.

The N2O5 reacted is O2, and the O2 produced by the N2O5 reaction is O2 equilibrium concentration, then the amount of O2 is 5mol L

Then the O2 produced by the N2O3 reaction is.

So N2O3 reacted, remaining.

So, the N2O3 equilibrium concentration is.