A chemistry calculation solution question for senior 1 and a chemical calculation problem for senior

1) Commentary: The heated gas must be the carbon dioxide released by NaHCO3, and the precipitation is calcium carbonate.

2）39，k

Narrator: n(nahco3)=

So m(naHCO3)=

So m(m2CO3)=

And because of v(gas)=

i.e. n(gas)=

n（na2co3）=n（co2）=

n(m2CO3) = it's n(CO2)=

Relative molecular weight of M2CO3.

Ask for M again. 3) The flame color reaction is very fast.

The original mixed solid is dissolved in water, a small amount of liquid is dipped in platinum wire, and then burned on an alcohol lamp, and the flame color is observed through the blue cobalt glass (because it contains the element Na, a yellow flame will be produced, and the blue cobalt glass can filter out the yellow).

Then you see a purple flame, which is unique to K (potassium).

1)2nahco3∽co2∽caco3xx=

2)2nahco3∽na2co3yy=

na2co3+h2so4=na2so4+co2↑+h2oz

z=v=calculated according to the molar volume of the gas)

m2co3+h2so4=m2so4+ co2↑+h202m+60 44

i.e. (m=39 is: potassium.

3) Use the flame color reaction, dissolve the original mixed solid in water, dip a small amount of liquid with a glass rod and burn it on the alcohol lamp, and observe the color through the blue cobalt glass (because it contains Na2CO3, the yellow light produced will cover the purple light of potassium, and the blue cobalt glass can filter out the yellow light.

(Mg2+ precipitate first) In the 0-10ml process, the quality of the precipitate remains unchanged, in the 10-30ml process, the precipitate.

The mass does not decrease but increases, indicating that Al(OH)3 is generated in this process, so there are NaOH and Naalo2 in the original solution.

There is no Al(OH)3, so, at 10 ml, the precipitate is all Mg(OH)2. The amount of its substance is , and the mass of MgCl2 is .

10-30ml process.

AlO2- +H+ +H2O = Al(OH)3N(Al3+)=20 1000*1mol=AlCl3 has a mass of .

At 30 ml, the solution becomes NaCl solution, N(Na+)=N(Cl-)=30 1000*1+=, so the mass of NaOH is.

When the precipitate mass is 0, n(cl-)=

v(hcl)=130ml

1.After adding 0-10ml of hydrochloric acid, the quality of the precipitate does not change, indicating that the hydrochloric acid is neutralizing sodium hydroxide at this time, so it means that the sodium hydroxide is excessive for the precipitation reaction of Alcl3, so when the volume of hydrochloric acid is added to 0-10ml, the precipitate is magnesium hydroxide.

2.Because it has been proved in the first question that the precipitate is completely magnesium hydroxide, you can find out what is the mass of magnesium chloride, in junior high school, you can use the relative atomic mass and relative molecular mass, in high school, you can use the knowledge of the quantity of matter, and the final answer should be.

In addition, from the first question, you can also get 20ml of hydrochloric acid for the generation of aluminum hydroxide precipitation, and at this time, the hydrochloric acid reflects the sodium metaaluminate that has been reacted with excess sodium hydroxide, according to the equation, the ratio of sodium metaaluminate and dilute hydrochloric acid can be obtained is 1:1, if you know the amount concentration of hydrochloric acid, you can find the amount of hydrochloric acid, so that the amount of aluminum ion can be found, and then the mass of aluminum chloride can also be found.

Or from the first question, I know that 10ml of dilute hydrochloric acid is needed to neutralize those sodium hydroxide, and sodium metaaluminate also has sodium ions, so if you want to find the mass of sodium hydroxide solids, you must add all of them together and then ask for it.

3.Because the amount (or mass) of the precipitated substance is already known according to the previous two questions, it can be found by adding the volume of hydrochloric acid required to neutralize sodium hydroxide, the volume of hydrochloric acid required to form the aluminum hydroxide precipitate, and the volume of hydrochloric acid when all the precipitates react with hydrochloric acid completely.

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.