Put 5.6 g of iron powder into 80 g of copper sulfate and react exactly. Calculate (1) the mass fraction of solute in the original copper sulfate solution (2) the mass fraction of solute in the solution after reaction

Put 5.6 g of iron powder into 80 g of copper sulfate and react exactly. Calculate (1) the mass fraction of solute in the original copper sulfate solution (2) the mass fraction of solute in the solution after reaction

Suppose the mass of solute in the original copper sulfate solution is x, the mass of Cu generated by the reaction is y, and the mass of FeSO4 generated is z.fe + CuSO4 = Cu + feso456 160 64 1525.6g x y z56 / 5.6 = 160 / x = 64 / y = 152 / ZX = 16gy = 6.4gz = 15.2g. The mass fraction of solute in the original copper sulfate solution is 16 / 80 * 100% = 20% of the solution after the reaction

Put the excess iron powder into the beaker of the mixed solution of dilute sulfuric acid and copper sulfate, filter after the reaction, and the solute contained in the filtrate is () A. Fe2（SO4）3 B. CuSO4 C. FeSO4 D. All three may exist

A. Iron can react with sulfuric acid to produce ferrous sulfate and hydrogen, and can react with copper sulfate to produce ferrous sulfate and copper, but can not produce ferric sulfate, so a is wrong;
B. Excessive iron powder can completely react with copper sulfate, so it will not contain copper sulfate, so B is wrong;
C. Iron can react with sulfuric acid to produce ferrous sulfate and hydrogen, and can react with copper sulfate to produce ferrous sulfate and copper, so C is correct;
D. Excessive iron powder can completely react with sulfuric acid, so it will not contain sulfuric acid, so D error
Therefore, C

(senior one chemistry) put excess iron powder into copper sulfate solution containing sulfuric acid for full reaction, and then filter the solution, The excess iron powder is put into the copper sulfate solution containing sulfuric acid for full reaction, and then the solution is filtered. The mass of filter residue is equal to that of iron powder. What is the ratio of sulfuric acid and copper sulfate in the original mixed solution?

The mass of filter residue is equal to the mass of iron powder, indicating that the mass of Fe reacted is equal to the mass of Cu formed
Fe+H2SO4=FeSO4+H2
y
CuSO4+Fe=FeSO4+Cu
x x x
Let the amount of iron reacting with sulfuric acid be ymol and the amount of iron reacting with copper sulfate be xmol
Then, (x + y) * 56 = 64y
y:x=1:7

The solution obtained by putting excess iron powder into copper sulfate solution and filtering after complete reaction is called

Fe+CuSO4===FeSO4+Cu
The solution obtained by filtration is called ferrous sulfate solution

5.6g iron reacts exactly with 100g copper sulfate, 1. In 100g copper sulfate solution, the chemical formula of the solute is __, Its quality is_______ g. The solvent is __, Its quality is_______ g. 2. After complete reaction, it is formed______ The chemical formula of solution and solute is_____ g. Quality is_____ g. The mass of the solvent is_____ g. The mass of the resulting solution is_____ g. CuSO4 16 water 84 FeSO4 15.2 84 99.2 How do you calculate these tens,

Fe+CuSO4=FeSO4+Cu
56 160 152 64
5.6 16 15.2 6.4
100-16=84 84+15.2=99.2

5.6g iron reacts exactly with 100g copper sulfate solution 1. In 100g copper sulfate solution, the chemical formula of the solute is __, Its quality is_______ g. The solvent is __, Its quality is_______ g. 2. After complete reaction, it is formed______ The chemical formula of solution and solute is_____ g. Quality is_____ g. The mass of the solvent is_____ g. The mass of the resulting solution is_____ g.

1. The mass of CuSO4 is 16g, and the mass of H2O is 84g2. The mass of FeSO4 in light green solution is 15.2g, the mass of solvent is 84g, and the mass of solution is 99.2g. Fe + CuSO4 = = FeSO4 + Cu 5.6gfe = 0.1mol 0.1 0.1 0.1, so the mass of copper sulfate is 0.1mol 0.1 * 160 = 16g, water 100-16 = 84g, and the mass of ferrous sulfate is 0.1