Question

Сколько граммов раствора нитрата серебра(AgNO3) с массовой долей соли 10% необходимо прилить к 80г раствора с массовой долей раствора хлорида бария(BaCl2) 5%, чтобы полностью удалить ионы серебра(Ag) из раствора.

Answer 1

МАССА хлорида бария=80х0,05=4 грамма
2AgNO3+BaCl2=2AgCl+Ba(NO3)2
молярная масса хлорида бария=137+71=208 г\моль
молярная масса AgNO3=108+14+48=170 г\моль
составляем пропорцию
 на 2х170 г нитрата серебра ---------------нужно      208 г хлорида бария
                Х-----------------------------------------4 ГРАММА
Х=2х170х4\208=6,54 г
по определению  концентрации в 100 граммах раствора ------------10 г нитрата серебра
                                                               Х----------------------------------------6,54
Х=100х6,54\10=65,4 г раствора нитрата серебра 10%-ного
                     

Answer 2

Calculation of the Amount of Silver Nitrate Solution Needed

To calculate the amount of silver nitrate solution (AgNO3) with a 10% mass fraction of salt needed to completely remove silver ions (Ag) from a solution containing 80g of barium chloride (BaCl2) with a 5% mass fraction of the solution, we can use the following steps:

1. Determine the mass of silver ions (Ag) present in the barium chloride solution: - The mass fraction of the barium chloride solution is given as 5%, which means that 5% of the solution's mass is barium chloride. - Therefore, the mass of barium chloride in the solution is 0.05 * 80g = 4g.

2. Calculate the molar mass of barium chloride (BaCl2): - The molar mass of barium (Ba) is 137.33 g/mol, and the molar mass of chlorine (Cl) is 35.45 g/mol. - The molar mass of barium chloride (BaCl2) is (1 * 137.33) + (2 * 35.45) = 208.23 g/mol.

3. Convert the mass of barium chloride to moles: - Divide the mass of barium chloride (4g) by the molar mass of barium chloride (208.23 g/mol) to get the number of moles of barium chloride: 4g / 208.23 g/mol = 0.0192 mol.

4. Determine the stoichiometric ratio between silver nitrate (AgNO3) and barium chloride (BaCl2): - The balanced chemical equation for the reaction between silver nitrate and barium chloride is: 2 AgNO3 + BaCl2 → 2 AgCl + Ba(NO3)2. - From the equation, we can see that 2 moles of silver nitrate react with 1 mole of barium chloride to form 2 moles of silver chloride. - Therefore, the stoichiometric ratio between silver nitrate and barium chloride is 2:1.

5. Calculate the moles of silver nitrate needed: - Since the stoichiometric ratio between silver nitrate and barium chloride is 2:1, we need twice the number of moles of barium chloride to react completely. - Multiply the number of moles of barium chloride (0.0192 mol) by 2 to get the moles of silver nitrate needed: 0.0192 mol * 2 = 0.0384 mol.

6. Convert the moles of silver nitrate to grams: - The molar mass of silver nitrate (AgNO3) is 169.87 g/mol. - Multiply the moles of silver nitrate (0.0384 mol) by the molar mass of silver nitrate to get the mass in grams: 0.0384 mol * 169.87 g/mol = 6.53 g.

Therefore, approximately 6.53 grams of silver nitrate solution with a 10% mass fraction of salt is needed to completely remove silver ions from the solution containing 80 grams of barium chloride with a 5% mass fraction of the solution.

Please note that the given search results did not provide specific information about the molar mass of silver nitrate solution or the concentration of the solution. Therefore, the calculation assumes that the 10% mass fraction of salt refers to the concentration of silver nitrate in the solution.