Continue to the chemistry questions of the first year of high school Please listen to the questions

Solution: Let the mass of Ca(OH)2 required for the reaction with 10g(NH4)2SO4 be X, and the amount of substance to form NH3 is N

nh4)2so4+ca(oh)2=2nh3 +2h2o+caso4132g 74g 2mol

10g x n

x =<10g .Therefore, Ca(OH)2 is excessive.

n(nh3) =2mol×10g/132g =v=n/vm=

c (ammonia) =

Answer: The amount of the substance concentration of ammonia is.

The chemical reactions that occur are:

ca（oh）2 + nh4)2so4 = caso4 + 2nh3 + 2h2o

10g 10g x

From the reaction, it can be seen that to consume 10gCa(Oh)2, more than 10g of (NH4)2SO4 is required, so Ca(Oh)2 is excessive. It needs to be calculated according to the quality of ammonium sulfate. Let the mass of the obtained ammonia be x, then.

132 10g = 34 x, solution x =

The amount of ammonia = , the volume under the standard condition =

The amount of the substance concentration of ammonia =

Calculated from 3Fe + 8H+ +2NO3 - 3Fe2+ +2No +4H2O:

1. "The decrease in the mass of iron powder indicates that the iron participating in the reaction is, and all the hydrogen ions are in nitric acid, so nitric acid is, and the concentration = mol l;

2. "The decrease in the mass of iron powder means that the iron has reacted again, and the hydrogen ion is in the sulfuric acid, for, so v= .

Reacted FE is for.

The amount of matter of the gaseous produced is.

Since there is still a residual Fe element after the reaction is complete, all the Fe2+ in the solution is Fe2+, that is, the transferred electrons are altogether.

The amount of unreduced Hno3 and Fe2+ forms Fe(No3)2, that is, the amount of nitric acid that has not been reduced is.

The amount of nitric acid reduced is the amount of the gaseous substance.

Therefore, the ratio of the amount of reduced and unreduced Hno3 species in the reaction is the original amount of substances with total Hno3 is.

i.e. at the concentration. Let no be xmol and no2 be ymol

Then the conservation of transfer electrons is 3x+y=

The conservation of the n element is x+y=

The solution is x=, y=

Therefore, the ratio of the volume of no to no2 is the ratio of the quantity of the substance is 3:1

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

AHGO will decompose into HG and O2 under heating conditions, and the oxygen generated by 2HGO===HG+O2 will react with the metal NA, and the reaction product of the reaction between NA and oxygen is determined by the ratio of the amount of NA to the oxygen substance

4na+o2====2na2o

na+o2==2nao；The condition of this question says that the composition of the air has not changed, that is, the oxygen in the air has not been consumed or increased, so there is no Na2O in the product of the reaction between sodium and oxygen, it is all NaO (otherwise it will consume the oxygen in the air), and the oxygen released by Hgo and Na just react completely to form NaO, so Na:O2=2:1,Na:

hgo=1:1

The correct answer is a

2hgo=2hg+o2↑

2na+o2=2na2o2

Total reaction: 2Na + 2Hgo = Na2O2 + 2Hg Therefore, the ratio of the amount of sodium and mercury oxide substances is 1:1

Choose a because Hgo will decompose into Hg and O2 under heating conditions, and the oxygen generated by 2Hgo==Hg+O2 will react with metal Na, and the reaction product of Na and oxygen reaction is determined by the ratio of the amount of Na to the oxygen substance: 4Na+O2==2Na2O; na+o2==2nao；The condition of this question says that the composition of the air has not changed, that is, the oxygen in the air has not been consumed or increased, so there is no Na2O in the product of the reaction between sodium and oxygen, it is all NaO (otherwise it will consume the oxygen in the air), and the oxygen released by Hgo and Na just react completely to form NaO, so Na:O2=2:

1，na：hgo=1:1

(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≈