Chemistry calculation questions for the first year of high school, chemistry calculation questions f

Question 11 n2---3h2*****==2nh3--- the amount of substance decreases.

x---y---1 + mol.

x=mol; y = moles.

The n2 participating in the reaction is molar. The amount of matter of H2 at equilibrium is moles.

h2 conversion rate:

12 questions. 2SO2---O2*****==2SO3--- gas volume decreases.

x---y---z---8+6)*10%==x==;y== ;z=

The conversion rate of SO2 is .

The volume of the gas mixture at equilibrium is 8+, and the volume fraction of SO3 is that you don't seem to know the difference method).

Trick Solution:

11 The total amount of gas decreases and the amount of matter is equal to the amount of NH3. Therefore, NH3 is. Hence the amount of N2 species participating in the reaction is. The amount of H2 in equilibrium is 3. αh2）＝

12．14×10％＝。Therefore, the conversion amount of O2 is.

so2）＝so2）＝（8－

2 h2s + 3 o2 = 2 h2o + 2 so2

60ml 90ml 60ml 60ml

So you get 60ml of sulfur dioxide and you get 10ml of hydrogen sulfide.

In fact, it is NACL and NABR that started to pretend to be Lu

In the end, it became NACL

It's that BR is replaced by CL.

The reduced mass of the solids is.

According to 1:1 for the early age of the stove generation.

br---cl---

x solution yields x=8g

The mass fraction of the original br is 8 20 = 40%.

Because the mass of the generated solution is equal to the mass of the original BaCl2 solution, the mass of the added sulfuric acid solution is equal to the mass of the barium sulfate precipitate generated, and 1 mol sulfuric acid participates in the reaction, so the mass fraction of the sulfuric acid solution is 98 233*100%=, so the concentration of its substance is: 1000*

Until the precipitation is exactly complete. It is known that the mass of the resulting solution is equal to the mass of the original barium chloride solution == indicates that the mass of the barium sulfate precipitate === the mass of the sulfuric acid solution.

H2SO4+BACl2*****===BASO4+2HCl98G---233G: The mass fraction of sulfuric acid solution is 98 233*100%==I hope it will help you (

H2SO4+BACL2=BASO4+2HCl, because the mass of the resulting solution is equal to the mass of the original BaCl2 solution, the mass of the precipitate generated is equal to the mass of the original sulfuric acid solution. If 233 grams of barium sulfate are generated, 98 grams of sulfuric acid are required, and the mass of the sulfuric acid solution is 233 grams.

Amount of sulfuric acid: 98 98 = 1

Volume: 233

c=1 mass fraction=98 233

Since the amount of CO2 is generated, and 1molco is formed from the reaction equation: CO+CuO=CO2+Cu, the amount of CO2 in the mixed gas is, m=mn=28

The mass of CO2 in the mixed gas is 18-14 = 4g, n = m m = 4 44 = 1 11mol, and the volume is 1 11

CO2 volume is, density p=m v=(

If you assume that Co is x grams, then CO2 is 18-x grams, and the molar mass of CO2 expressed as X is calculated according to the formula Co+CuO=Cu+CO2. That's it.