A chemistry calculation problem A chemistry calculation problem

The disinfection principle of CLO2 and Cl2 is to use their strong oxidation, from the perspective of the disinfection process, these two substances must be converted into HCL to get 5 electrons, and Cl2 to HCl to get 2 electrons. Both substances are one mole, and the disinfection efficiency of ClO2 is twice that of Cl2. If it is calculated in terms of unit mass:

Specifically: [(1

The end result is multiple.

8clo2+4h2o=3hcl+5hclo4

clo2---cl-

There is no way to write an equation, you can give an equation like this: clo2+h2o---3[o]+2h+ +cl- (there is no way to balance).

5mol electrons are obtained each.

5 electrons per gram are obtained.

cl2---2cl-

Every 2mol of electrons is obtained.

2 71 = electrons per gram are obtained.

The disinfection efficiency of ClO2 is twice that of Cl2.

In the first case, Fe(No3)3 is all generated, HNo3 is 36, and the reduced nitric acid is.

In the second case, only Fe(No3)2 is generated, HNo3 is 24, and Hno3= is reduced

In the third case, Fe(No3)2, Fe(No3)3, 24 Hno3 36 are generated, and Hno3 is reduced

2)b＞13a

3) The larger, the more C (

The main thing is to grasp the number of electrons per unit mass of oxidant to calculate:

These two substances are disinfected by using their oxidation, and they are reduced from ** to negative monovalent.

clo2 --- to -1 valence (electrons are 5).Its unit mass is grams.

Cl2 works in disinfection with HClo--to-1 valence (get 2).Its unit mass is 71 grams.

Then:[(1

Write out the chemical equation and calculate.

Solution: Let the number of crystal water in the molecular formula of ferrous sulfate crystals be x and the mass of the solution decreased after the crystals are precipitated.

The mass of solute reduction after precipitation of crystals is.

Hence the x 7

Solution: Let the value of the crystal water in the ferrous sulfate crystal be x.

The mass of water in the filtrate is m1 = (500 + 40 — g = g The mass of water in the precipitated crystal m2 = 500 * 3 5g - The mass fraction of water in the crystal is: 18x (152 + 18x) = x = 7 of the solution equation Answer:

This problem is quite simple, first of all, from the overall point of view, the mass of the remaining solution is, because the temperature has not changed, the mass ratio of solute to solvent in the solution has not changed, it is easy to get, the mass of Feso4 in the solution is, the mass of water is, so that according to the conservation of mass, the mass of FeSO4 of the precipitated crystal is 500*2 5+, and the mass of water is, so that the value of x is obtained according to the ratio of Feso4 and H2O molecules, and the calculation formula is:, get, x=7

Determine that the first is magnesium hydroxide, (aluminum hydroxide and sodium hydroxide reaction) mg is When hydrochloric acid is added, CO2 begins to be produced, and when hydrochloric acid is added, it is exactly the reflection that the hydrochloric acid that reacts with carbonate is 8:1 The number of hydroxide is the number of carbonate According to the conservation of electricity, it can be concluded that AL is integrated to obtain AL(2)MG(6)OH(16)CO3 What should I do if there is a little defect Add water molecules to make up.

al(2)mg(6)oh(16)co3 .H2O(4) altogether, and that's it.

PS: The hydrochloric acid and carbonate that start the reaction first produce bicarbonate and then CO2).

1) The maximum concentration of the amount of magnesium ions:

2) The amount of substances with sulfate ions does not exceed:

1) Maximum: mg2+ molar mass: 24g mol= So the maximum.

2) The quality of the sulfate ions.

Go to the amount of substance =

This question is the amount of potassium sulfate substance concentration just to confuse you.

It can be known that the amount of substances required for sulfuric acid is: (mol).

Then the concentration of sulfuric acid is: (1000* (mol l) (formula: c = 1000 pw m, p: density, w: mass fraction, m: relative molecular mass).

The volume to be taken is: (l).

(1) The amount and number of moles of Na2CO3 substance is the same as the number of moles of emitted gas, that is, 1mol of Na2CO3 is equivalent to 1120ml of 1mol, that is, the gas under standard conditions.

Na2CO3*(XH2O) is.

2) From the chemical equation, it can also be concluded that HCl is, a total of 10ml, and the concentration of dilute hydrochloric acid is 10mol L.

3) The molar mass of sodium carbonate (XH2O) is 106+18x, and the molar number is the first question).

The mass is (106+18x)*

Launch x = 10

The chemical reaction equation is attached: Na2CO3 + 2HCl = 2NaCl + H2O + CO2 (gas).

It should also be noted that because hydrochloric acid is added dropwise dropwise, it is an excess of Na2CO3 at the beginning, which will first react with 1molHCl to form NaHCO3 and NaCl, but no gas is produced, and when Na2CO3 is fully reacted to become NaHCO3, then HCl is added dropwise to produce CO2. Since it is written in the title that there is gas production, it is judged that the final HCl is excessive. All Na2CO3 is converted to NaCl and CO2.

Gas CO2, 1120ml is that is, moles.

According to the conservation of sodium carbonate (XH2O) is also moles.

x=(286-106)/18=10

na2co3---2hcl--co2

HCL is molar concentration=

(1)(2)10mol/l

1) Because 1120ml of gas (carbon dioxide) is collected, i.e. the carbon contained in sodium carbonate.

Because 1molNa2CO3 contains 1molc in it

So there is na2CO3 in it

That is, there is Na2CO3·XH2O

2) Because only sodium carbonate reacts with dilute hydrochloric acid.

So sodium carbonate requires hydrochloric acid.

Because exactly 10ml of hydrochloric acid is used, which contains hydrochloric acid, the hydrochloric acid concentration is.

3) Because there is Na2CO3

Therefore, there is water in it, i.e., water in hydrated sodium carbonate 9 18 = i.e., 1 mol of sodium carbonate and 10 mol of water in it

That is, the original substance is Na2CO3·10H2O

2CO2 + 2Na2O2 ==2Na2CO3 + O22H2O + 2Na2O2 ==4NaOH + O2 There is the equation on page 12 of the Lujiao textbook).

Let the hydrocarbon be cxhyThen:

CXHy+(y 4+x) O2===xCo2+y 2 H2O, so we get: 20L 14 4L+(X 2+4 Y)*Solvable to obtain: x=

And because of the increased mass of 15 grams of CO2 and H2O O2, so: 44x+y 2*18-x 2*32-4 y*32=15 solution y=1mol

So the original hydrocarbon is:

c5h10...

After seeing the upstairs, he was a little ashamed, his method was not bad, but he didn't understand it clearly. I'm writing a bit messy, and there's something I don't understand

The products of the complete combustion of hydrocarbons are carbon dioxide and water.

CO2 + Na2O2 ==Na2CO3 + Weight gain = 44-16 = 28

H2O + Na2O2 ==2NaOH + Weight Gain = 18-16 = 2 Weight Gain As long as you consider what comes out: in the first equation, the mass of carbon dioxide enters the solid, and the mass of oxygen leaves the solid, so weight gain = carbon dioxide minus oxygen].

In the end, all that is left is oxygen, note that it is under 10 °C, conditions, to convert to 0 °C, conditions, that is, the standard state, just divide by (273 + 10) and multiply by 273. The sum of the moles of water and carbon dioxide can then be obtained. It is obtained by using the total weight gain = 15 grams of simultaneous equations.

mg and naOH do not react.

And 2Al+2NaOH+2H2O=2Naalo2+3H2 then the mass of MG is CG

The quality of Al is A-CG

The volume of hydrogen produced is BML=B 1000L

Then the amount of matter is (b 1000).

Let the relative atomic weight of al be m

Then the amount of Al is (a-c) m

According to the proportional formula, the relationship between the amount of Al and the amount of H2 is 2:3=(a-c) m:b 22400

The solution yields m=33600(a-c) b

Only aluminum reacts with sodium hydroxide.

2al---3h2

x (b*10^-3/ mol x=(b*10^-3/ mol

Molar mass of aluminium = (a-c) (b*10 -3 Relative atomic mass of aluminium = (a-c) (b*10 -3

n(al)=3 2n(h2)=( Conservation of electrons gained and lost 1molal loses 3mol electrons, generated.

m(al)=a-c

m(al)=m(al) n(al)=33600(a-c) 3b or more units are omitted.