Two chemistry calculation problems, trouble master!

1.Solution: The amount of copper sulfate in the original copper sulfate solution =

Let the amount of substances involved in the reaction of copper sulphate be n

fe+cuso4=cu+feso4

56g...64g...The amount of solid increase in 1mol copper sulfate reaction m = 8g8g * n =

n = the amount of the substance remaining copper sulphate =

The amount concentration of the substance of copper sulphate after the reaction c (CuSO4) = A: The amount concentration of the substance of copper sulphate after the reaction is.

2.Solution: The chemical formula of sodium sulfate crystals is.

The amount of the substance that consumes barium chloride n (BaCl2) = 5mol L * Let the amount of the substance of sodium sulfate be n.

na2so4+bacl2=baso4+2nacln...

n = molar mass of sodium sulfate crystals m (

then 46 + 96 + 18x = 322

The chemical formula of x=10 sodium sulfate crystals is:

Answer: The chemical formula of sodium sulfate crystals is:

1. The amount of copper sulfate in the reaction is xmol

cuso4+fe=feso4+cu

1 mass increase 64 56 8xx

The concentration of the amount of copper sulfate species in the solution after the reaction = (

2. 142+18n 1

The chemical formula of n=10 sodium sulfate crystals is:

Question 1:

Gram. Question 2: Write the equation (slightly trembling) to produce a gas of 165 + 53-196 = 22 grams, and if the mass fraction is or ruler w is w, then the caco3 participating in the reaction is 53w

Columnar. 100/53w=44/22

w=let's generate cacl2

x grams, then 111 x = 44 22

x = grams. So the quality fraction is:

Throw away the minutiae and see the main reaction processes and metrological relationships. This is a question about iodine dosing. Potassium iodate (indicated by a) reacts with potassium iodide (indicated by b) to form elemental iodine (c); Sodium thiosulfate (d) reacts with iodine to form iodine ions (e) and tetrasulfate (f).

Okay, let's start solving the problem, I will directly use the conservation of electron transfer to list all the measurement relations (I won't write about the water god horse in the specific equation, trouble), a+5b=3c; 2d+c=2e+f；Then using these two relationships, the amount of d consumed can be solved. That is, you first calculate the amount of C generated, and then calculate the amount of D consumed, and since you know the volume, you can calculate the concentration of the sodium thiosulfate standard solution. There's a lot of data in this that you can't use when calculating, which are just some auxiliary parameters for experiments.

Uh, is that detailed enough?

（1）n（na+）=8 mol•l—1× mol

n(so 42-)=3 mol l—1 mol (1 point).

If the precipitate is only Mg(OH)2, there are only Na+, AlO2- and SO42- in the solution, and it is electrically neutral.

then n(na+) = 2n(so42-)n(alo2-)2 points).

n（al）=n（alo2- ）= mol

m（al）= g×27 g•mol－1= g

m(mg) = g g = g (1 point).

then 60% mg) 100% (2 points).

5 points) (2) When m(al), m(sin) = 58 [24 (2 points).]

When, m(shen) = 78 [m(al)27 + 58[(2 points)].

Shen) 2 points).

This is an adapted question, and the original question is hard to find.

The equation for the conservation of electrons n(fe3+)*1=n(i-)=0,02mol

n(fe)= then n(fe2+)=

According to the conservation of electrons of gain and loss, the equation of the element conservation column n(hno3)=n(no)*3=n(fe2+)*2+(fe3+)*3=, c(hno3)=