 # 249.5 g of dilute sulfuric acid is added to a beaker containing 50 g of calcium carbonate and the reaction is just complete. Try to calculate the mass fraction of solute in the solution obtained after the reaction

## 249.5 g of dilute sulfuric acid is added to a beaker containing 50 g of calcium carbonate and the reaction is just complete. Try to calculate the mass fraction of solute in the solution obtained after the reaction

Let the mass of generated calcium chloride be x and the mass of carbon dioxide be y
CaCO3+2HCl==CaCl2+H2O+CO2↑
100 111 44
50g X Y
100:111=50g:X
X=55.5g
100:44=50g:Y
Y=22g
Mass fraction of solution after reaction = 55.5g / (249.5g + 50g-22g) = 20%

### The mass fraction of solute in the product solution of 10G calcium carbonate and 100g dilute HCl 10g calcium carbonate reacts exactly with 100g dilute HCl. Calculate the mass fraction of solute in the product solution

1. According to the chemical equation, 4.4g of carbon dioxide can be obtained from 10g of calcium carbonate. The mass of the solute calcium chloride after the reaction is 11.1g. 2. According to the law of mass conservation, the total mass of the reactant is equal to the total mass of the product. It can be seen that the mass of the solution after the reaction is 10g + 100g-4.4g = 105.6g3. According to the mass fraction of the solute

### 50g hydrochloric acid solution with a certain concentration reacts exactly with 10g CaCO3 powder, and the mass fraction of solute in the solution is calculated?

CaCO3 + 2HCl = CaCl2 + CO2 + h2o100 ------ 73 --- 111 -- 4410 ------ X1 --- x2 --- x310 / 100 = X1 / 73 = x2 / 111 = X3 / 44. The solution shows that X1 = 7.3, X2 = 11.1, X3 = 4.4. The solute in the solution is only calcium chloride, and the mass is 11.1g. The total mass of the solution = 50 + 10-4.4 = 55.6g

### Put 20g calcium carbonate powder into a certain amount of hydrochloric acid and react exactly completely. The mass fraction of solute in the obtained solution is 20%. Try to (1) generate carbon dioxide Put 20g calcium carbonate powder into a certain amount of hydrochloric acid and react exactly completely. The mass fraction of solute in the obtained solution is 20% Try to find (1) how many grams of carbon dioxide? (2) What is the mass of the solution obtained after the reaction? (3) What is the mass fraction of solute in the original hydrochloric acid solution?

CaCO3+2HCl=CaCl2+H2O+CO2100 73 111 4420g y z X100/20g=44/xx=8.8g 100/20g=73/yy=14.6g100/20g=111/zz=22.2g22.2/20%=111g20g+14.6g-8.8g=25.8g111g-25.8g=85.2g14.6g/(14.6+85.2)=14.6%(1)8.8g(2)111g(3)14.6%

### 100g CaCO3 reacts with HCl to produce 8.8g CO2. Calculate: 1. Mass fraction of calcium in the sample. 2. Mass fraction of solute in the final solution

CaCO3---------CO2
20---100 8.8---44
Calcium carbonate has 20g
Mass fraction of calcium: (40 / 100) * 20 / 100 = 8%
The content of HCl in the solution is unknown, and the water is unknown, so the mass fraction cannot be obtained

### Add 50.3 g of dilute hydrochloric acid to a certain amount of calcium carbonate and sodium chloride, and the reaction is just complete to obtain 4 and 4 g of CO2, and then add 20 g of CO2 to the resulting solution Add 50.3g dilute hydrochloric acid to a certain amount of calcium carbonate and sodium chloride, and the reaction is just complete to obtain 4.4g CO2. Then add 200g AgNO3 solution (sufficient amount) to the resulting solution to obtain 86.1g white precipitation. Calculate the mass fraction of calcium ion in the solution after reaction

4.4 g CO2 is 0.1mol, so calcium ion is 0.1mol because it happens to react
The white precipitate is AgCl, 0.6mol Cl ion comes from HCl and NaCl
After removing 0.2mol of Cl ion from CaCl2, 0.4mol of NaCl is 58.5 * 0.4 = 23.4g
CaCO3 is 0.1mol and the mass is 10g
N (Ca +) = 0.1mol, mass 4G
According to the mass conservation, the total mass of the solution is 10 + 23.4 + 50.3 + 200-4.4-86.1 = 193.2g
4/193.2=2%
A little hasty, I don't know if it's right