Chemistry questions, chemistry questions, please answer in detail

1. Potassium hydroxide.

2、n(hcl)=(

Carbon dioxide, mass, can be calculated using chemical equations.

3. v (carbon dioxide) =

Its volume fraction =

There is also potassium hydroxide, and the following idea is to multiply by 5 percent and divide by the non-mass concentration of hydrochloric acid, and then find the amount of carbon dioxide according to the chemical equation. Finally, divide the amount of carbon dioxide by the density.

1 Potassium bicarbonate.

2 Volume of carbon dioxide v1=

CO2 Volume Fraction =

1), and potassium hydroxide;

hcl + k2co3 = 2kcl + h2o + co273g 138g 44g

Solving the equation by x y shows that x=; y= ；

koh + co2 = k2co3 + h2o44g 138g

The z-solution equation shows that z=;

v(co2)=；

w(co2)= 5=;

Can react. This is a reaction of strong acids to weak acids to produce hexametametaphosphoric acid.

napo3)6 + 6hcl = 6nacl + hpo3)6

1）2hbr+ci2==2hci+br2

h2so3+ci2+h2o==h2so4+2hci

h2so3+br2+h2o==h2so4+2hbr

The law of balance can be played to the fullest here. Let's assume that the amount of HBR and H2SO3 is X Mo.

Lost electrons: x 2 mos.

HBRX MoH2SO3Total loss of 2x+1 2x mo.

Gain electron: Moc2 lost Mo.

So there is 2x+1 2x=

Solution x== Mo.

The amount concentration of HBR and H2SO3 is C = Molliters.

2）2co+o2==2co2---1）

2h2+o2==2h2o---2）

1) From the above, the mixture gas is mixed in any ratio, and the volume of O2 required for complete combustion is 5L (half of the mixed gas), then the remaining O2 is 10L, and the remaining 15L is the CO2 generated (water has been removed).According to equation (1), it is not difficult to conclude that the original mixture contains a volume ratio of carbon monoxide to hydrogen that is 1:1

2) After the reaction, the remaining CO2 is (a-10)lThe CO in the original mixture is (A-10)L

h2 is (20-a)lIt turns out that the volume ratio of carbon monoxide to hydrogen is (a-10) (20-a).

:(a) is the hydrolytic equilibrium and (b) is the ionization equilibrium.

2) the ionization equilibrium shifts to the right; The hydrolytic equilibrium shifts to the left; c(h+) becomes larger.

3) C(S2) becomes larger.

4) Large.

A small amount of SO2 reacts with NaOH in small amounts and in excess of it, respectively.

Sulfur dioxide, which corresponds to sulfurous acid, is a dibasic acid that can react with alkali to form positive salts or acid salts. Specifically, when there is an excess of alkali (a small amount of sulfur dioxide), positive salts are generated, and when there is an excess of acid (a small amount of alkali), acid salts are generated.

Since sulfurous acid is stronger than carbonic acid, SO2 reacts with NaHCO3 to form sodium sulfite, carbon dioxide, and water.

Adopt it... Hehe.

A small amount of SO2 is used to generate Na2SO3

Excess SO2, NaHSO3

CO2 because sulfurous acid is more acidic than carbonic acid.

Sulfur dioxide reacts with a small amount of sodium hydroxide to form Na2SO3, and with excess NaOH to form NaHSO3, because this is the reaction of an acidic oxide with a alkaline solution and the acid corresponding to SO2 is a dibasic weak acid.

SO2 reacts with NaHCO3 to form NA2SO3, H2O, CO2 because H2SO3 is more acidic than H2CO3.