Urgent high school chemistry equation all including small oh arithmetic, sulfuric acid, aluminum, et

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fe + s = fes (1)

The sum of the equations Fes + H2S04 = FeS04 + H2S (2)(1)(2) gives Fe + S + H2S04 = FeS04 + H2S

The amount of the substance emitting the gas is 3 8mol, and if the emitted gas is only H2S, the mass of the mixture should be 33g instead of 29g. Then rewind back and push, the condition holds, so FE is excessive.

If you look at the equation carefully, a molecule of iron reacts to form a molecule of gas (hydrogen or hydrogen sulfide), so the amount of iron can be determined based on the total amount of gas.

In fact, it can be understood that iron and acid react to form hydrogen, and part of hydrogen reacts with sulfur to form hydrogen sulfide. In this way, the amount of gas produced is only related to the amount of iron, and the amount of iron can be calculated according to the amount of gas generated, and then the amount of sulfur can be calculated, and then the amount of iron produced can be compared to determine whether there is too much iron.

Think about it yourself.

The appearance of the third proves that the feg is overdosed.

ch3CHO+O2====CH3COOHCH3CHh+H2====CH3CH2O2CH3CH2CH2OH+O2==Cu AG==2CH3CH2CH2CHo+2H2O2CH3OH+O2==AG CU==2CH2O+2H2O2CH3OH+2NA====2CH3ONA==H2CH2=CH2+H2O== (certain conditions)==ch3ch2ohch3cooh+ch3oh==(concentrated H2SO4, heated, reversible)==CH3cooch3+H2och3cooh+ch3ch2CH2OH== (concentrated H2SO4, heated, reversible)==CH3cooch2ch2ch3+H2och3cooh+ch3-choh-ch3==(concentrated H2SO4, heated, reversible)==CH3cooch(CH3)CH3+H2OCH2OOH+CH3OH==(concentrated H2SO4, heated, reversible)==CH2OOCH3+H2OOH+CH3CH2OH==(concentrated H2SO4, heated, reversible)==CH2OOCH2CH3+H2OOH+CH3CHHCH3==(concentrated H2SO4, heated, reversible)==CH2OOCH(CH3)CH3+H2OCH3CH2COOH+CH3OH==( Concentrated H2SO4, heated, reversible) ==CH3CH2COOCH3+H2OC3CH2COOH+CH3CH2OH==(Concentrated H2SO4, heated, reversible)==CH3CH2COOCH2CH3+H2OC3COOH+CH3CH2CH2OH==(Concentrated H2SO4, heated, reversible)==CH3COOCH2CH2CH3+H2OH (Is your teacher sick, this kind of question) o-o-h||ch3o2；h-c-h；ch2oohhho-o-h|||ch3h6o2；h-c-c-c-hch3ch2cooh||hhch3ch2cooch2ch2ch3+h2o==(alkaline conditions, heating)==ch3ch2cooh+ch3ch2ch2oh(c6h5)-cooch2ch3==(alkaline conditions, heating)==(c6h5)-cooh+ch3ch2oh tired old man, if it weren't for my good mood, I wouldn't write to you ==

The nitrification reaction of glycerol also needs to be heated.

1.Due to the different dropping sequences, the phenomena are different:

HCl and Na2CO3

HCl is added to Na2CO3: HCl + Na2CO3 = NaHCO3 + NaCl without gas.

Na2CO3 is added to HCl: 2HCl+Na2CO3=2NaCl+CO2+H2O drops into the gas immediately.

ALCL3 and NaOH

AlCl3 is added to NaOH: Al3++4OH-=ALO2-+2H20 without precipitate.

3alo2-+al3++6h2o=4al(oh)3 and then there is a precipitate.

NaOH is added to AlCl3: AlCl3 + 3NaOH = Al(OH)3 + 3NaCl has a precipitate to form.

Al(OH)3 + NaOH = Naalo2 + 2H2O precipitate dissolves.

HCL and Naalo2

HCl is added to Naalo2: H++ALO2-+H20=AL(0H)3 Existing precipitate Continue to add HCl and the precipitate disappears.

Naalo2 is added to HCl: ALO2-+4H+=Al3++2H20

agNO3 and NH3·H2O

AgNO3 drops are added to ammonia to generate a white precipitate agoh

The addition of ammonia to AGNO3 is a phenomenon in the silver mirror reaction, and the existing precipitation disappears after excess.

2. The phenomenon is different depending on the amount of dropping

na2co3+hcl=nahco3+nacl

na2co3+2hcl=co2+2nacl+h2o

Carbon dioxide passes into the clarified lime water, and there is a white precipitate (calcium carbonate) first, and when the carbon dioxide is excessive, the solution becomes clear again, and the product is calcium bicarbonate.

Ca2+ +2HCO3- +2OH- =CaCO3 +CO3(2-) 2H2O

co2 + clo- +h2o ==hco3- +hclo

cl2 + 2oh- =cl- +clo- +h2o

Mn in KMno4 decreases from +7 to +2, and S rises from -2 to 0 in H2S, and the number of gained and lost electrons is equal.

So the coefficient kmno4:h2s=2:5

The other coefficients are then trimmed. Pick D

The answer is that the price of two mns in D is reduced by a total of 10, so there should be five sulfurs oxidized.

d, the gains and losses of electrons should be conserved

d is chosen because the gain and loss electrons are not conserved.