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Let's analyze it first. Sodium reacts with water to form sodium hydroxide with hydrogen gas. Sodium hydroxide reacts with aluminum to form sodium metaaluminate and hydrogen.

When aluminum chloride is added, metaaluminate ions react with aluminum ions to form aluminum hydroxide precipitates. At 20 ml, the precipitation begins to precipitate, indicating that the hydroxide ions are excessive, and the added aluminum ions will directly react with the hydroxide ions to form metaaluminate ions. Until the hydroxide ion reaction is completed, it will be the aluminum ion that reacts with the metaaluminate ion to form a precipitate.

When a total of 40 ml is added, the precipitated precipitate reaches the maximum value, indicating that all metaaluminate ions in the solution have reacted at this time. And from the reaction equation, the amount of precipitated material is equal to the amount of aluminum ions added (40ml-20ml) 4 times. Let's start solving the problem.

n(al(oh)3)=4*20/1000*。When the precipitation reaches the maximum amount, then sodium chloride is left in the solution. From the conservation of elements, n(na)=n(cl-)=40 1000*, m(na)=.

Hydrogen: Sodium and water react to form hydrogen and sodium hydroxide, and aluminum and sodium hydroxide react to form hydrogen.

The 20 ml between 20 ml and 40 ml is the AlCl3 required for the reaction of AlCl3 with Naalo2 to form Al(OH)3, and the maximum amount of precipitated material can be calculated from this data.

When the precipitate reaches the maximum, the remaining ions in the solution are Na+ and Cl-, and according to the conservation of charge, the concentration of Cl- is calculated, that is, the concentration of Na+, from which the mass of the Na block can be obtained.

1、fe2o3 + 6hcl = fecl3 + 3h2ox 200 * 3 / 1000

Solve x = n(fe2o3) =

2. Add KSCN solution, there is no obvious phenomenon, indicating that there is no Fe3+, and all of them have been converted into Fe2+

2fe3+ +fe = 3fe2+

2 x solves x = n(fe) =

Then add iron powder at least m(fe) =

1)fe2o3+6hcl=2fecl3+3h2ox y

The amount of a substance of x= y=fe2o3 is.

2) At this time, KSCN solution was added dropwise to the solution, and there was no obvious phenomenon.

So the solution does not contain 3-valent Fe ions.

2fecl3+fe=3fecl2

zz = the mass of iron powder added is at least.

1.First trim chemical equation:

Fe2O3 + 6 HCl --2 FeCl3 + 3 H2OFE2O3 = 200 1000*3 6 = mol So the amount of Fe2O3 is mol

2 FeCl3 + Fe --3 FeCl2Fe2O3 is mol, so the amount of FeCl3 is mol so the amount of FeCl3 added = FeCl3 2 = mol iron powder mass = * 56 = grams.

1) Fe2O3 mol chemical equation dissolves exactly 1:6

2) FE chemical equation KSCN no phenomenon no FE3+ 2:1:

(1) Let the Cu atom have 5mol and the O atom have 3molCu with a mass of 64 5 = 320g

The mass of the o is 16 3 = 48g

m（cu）：m（o）=20：3

cu%=2) set cuoxmol, cu oymol

Yes (x+2y) (x+y) = 5:3

The solution is y=2x

That is, the ratio of the amount of matter of cuo and cu2o is 1:2

1:2 solution: Since there is only one o in both cuo and cu2o, it can be assumed that the cu atom is 5a mol and o is 3A MOL

So the mass fraction of the copper atom is 5a*64 (3a*16+5a*64)*100%=

Assuming that the ratio of cuo to cu2o is x, then cu2on mol can be assumed additionally, cuo is xn mol, so:

xn+2n) (xn+n)=5 3 to get x= so the ratio is 1:2

(1) The ratio of the quantities of known matter, which can be seen as five cu atoms and three o atoms. Therefore, the CU mass fraction (5*64) (5*64+3*16) is reduced to a percentage.

2) Let Cuo, the amount of Cu2O substances are n1, n2 (n1 + 2 n2) :(n1 + n2) = 5:3 The result I don't count, the landlord won't give points because of this, right? Hehe.

Solution: Let the quantity of cuo be x mol and the amount of cu2o be y mol, then according to the title, we can know:

x+2y)∕(x+y)=5/3

Calculated: y=2x

That is, the ratio of the amount of cuo and cu2o in the mixture is 1:2, then the mass fraction of the copper atoms in the mixture is:

w%=64(x+2y)∕[64(x+2y)+16(x+y)]=320∕368≈

The diagram is drawn on the computer, and you write the equation yourself.

The total solids added are 29g at the end, indicating that Reaction 2 binds more water than Reaction 1)

Then cao is more than naoh And then you can calculate all the answers by setting equations, and I can't play grass with a computer, so you can calculate it yourself One of the most important information to pay attention to in this question is"Ca2+, CO32- and HCO3- were no longer detectable in the solution"It means that the reactions in the solution are just completely reacted.

Question 2: From the ratio of the number of protons to the number of neutrons is 11:13, it can be seen that the mass of the element is a multiple of 24, and the relative molecular mass of the element is a multiple of 29, which is about multiplied by 48, so it is titanium, and the atomic mass of halogen is 142, and 142 is 4 times that of chlorine.

So the chemical formula is period 4, group ivb in the periodic table.

Let's analyze it, bauxite contains 60% of alumina, and the utilization rate of bauxite is 90%, that is to say, there is 90%*60%.

The total conversion rate of the reactants is 80%, that is to say, the reactants are not fully involved in the reaction, and 90%*60%*80% participate in the reaction

The rest is the reaction to make aluminum.

al2o3 ~ 2 al

x*90%*60%*80% 5*99%x=

5x99%=

27x2+16x3=102

27x2/102=

5x99%/(60%

I haven't done it for a long time, I don't know if it's right.

Using the conservation of aluminum is very simple.

Study hard.

Standard condition) - mol

molcl2 + 2 fe2+ =2fe3+ +2cl-x

x= (i.e., the elemental iron of the reaction and the iron in Fe2O3 are mol) with elemental iron a mol, Fe2O3 b mol is conserved by valence rise and fall (because the final solution composition is Fe(NO3)2 A elemental iron rises 2, a Fe2O3 decreases 2, and a NO3-falls 3).

So the column equation 2a=2b+3*valence rise and fall conservation) a+2b=atomic conservation)

The solution is a= b=

1) It turns out that the mass of the unoxidized iron in this rusty iron sheet is (2) because there is still 3g of elemental iron left, so the nitric acid completely reacts Fe-Fe(NO3)2

fe2o3——2fe(no3)2

The amount of no3- substance is mol

The amount concentration of the substance of the original nitric acid is mol = mol l

Let the fe that is reacted with x g (the unreacted fe is the fe that is taken out), then the sum of the electrons obtained by no and cl2 is equal to the electrons lost by fe, i.e., + (3 *x because in the end fe is oxidized to +3 by cl2) x = mol

So it turns out that the unoxidized iron has a total of 56*x+ g

After the reaction with nitric acid, the solute in the solution is only Fe(NO3)2, because there is still Fe left, and there will be no +3 valence Fe

So N(Hno3)=2N[Fe(No3)2]+N(No) and N[Fe(No3)2]=X= mol, so N(Hno3)=2* mol, so C(Hno3)= mol L

Amount of iron = (40 160) *2 = mol Amount of sulfur = = mol

Let Fes have x mol and Fes2 have y mol, and the equation :

x + y =

x + 2y =

The solution is x = y =

The quantity ratio of matter is 3:2