 # In 196 g sulfuric acid solution, 32.5 g zinc is added to just complete the reaction. Find the mass of hydrogen produced after the reaction?

## In 196 g sulfuric acid solution, 32.5 g zinc is added to just complete the reaction. Find the mass of hydrogen produced after the reaction?

Because it just reflects
So Zn is all used up
Zn+H2SO4=H2+ZnSO4
And Zn has 0.5mol
Therefore, H2 0.5mol was generated
The mass is 1g

### To 196 g of sulfuric acid solution, 32.5 g of zinc sheet were added·····

Zn + H2SO4=ZnSO4+H2
65 98 2
32.5g x y
x=49g y=1g
Generate 1 g of hydrogen
The solute mass fraction of sulfuric acid is
49g/196g*100%=25%

### 32.5g zinc particles react exactly with 280g sulfuric acid solution. Find (1) how many grams of hydrogen are generated? (2) What is the mass fraction of the solute in the sulfuric acid solution used? (3) What is the mass fraction of solute in the product solution?

Zn + H2SO4 = ZnSO4 + H2 (gas)
65 98 161 2
32.5g y z x
Zinc reacts completely when sulfuric acid is excessive
65/2=32.5g/x
x=1g;
65/98= 32.5g/y;
y=49g;
65/161=32.5g/z;
z=80.5g
(1) . the mass of hydrogen generated is 1g;
(2) The mass fraction of solute in the sulfuric acid solution used is: 49G / 280g * 100% = 17.5%;
(3) The mass fraction of solute in the product solution is 80.5g / (32.5g + 280g-1g) * 100% = 25.8%

### For hydrogen production, he added 32.5 g of zinc particles into 200 g of sulfuric acid solution to exactly complete the reaction, and calculated the mass fraction of solute in dilute sulfuric acid I want the results? fast

24.5%
Trust me. O (∩) o

### What is the mass fraction of solute in the solution obtained by the complete reaction of a 20g copper zinc alloy with 49G 20% dilute sulfuric acid?

Let the mass of zinc in the alloy be x, the mass of zinc sulfate generated by reaction be y, and the mass of hydrogen generated be Z
Mass of sulfuric acid in 49 grams of 20% dilute sulfuric acid = 49G * 20% = 9.8g
Zn + H2SO4 = ZnSO4 + H2↑
65　　98　　　　161　2
X 9.8g Y Z
X=6.5g
Y=16.1g
Z=0.2g
Solute mass in the obtained solution = 6.5g + 49g-0.2g = 55.3g
Mass fraction of solute in the obtained solution = 16.1g/55.3g * 100% = 29.1%
A: the mass fraction of solute in the obtained solution is 29.1%

### 26G zinc reacts with 206G dilute sulfuric acid to calculate the solute mass fraction of the original dilute sulfuric acid. The solute mass fraction in the solution after the reaction

26G zinc = 0.4mol, 206G dilute sulfuric acid solute should also be 0.4mol = 39.2g,
Solute mass fraction of original dilute sulfuric acid = 39.2 / 206 = 19.03%
The mass of solution in the solution after reaction is 206 + 26-0.4 * 2 = 231.2g
Mass of solute zinc sulfate: 0.4 * 161 = 64.4g
Mass fraction of solute in solution after reaction = 64.4 / 231.2 = 27.85%