Question

К раствор хлорида кальция содержащий 11,1 г соли прилили раствор содержащий 1,7 г нитрата серебра найти массу осадка

Answer 1

CaCl2+2AgNO3=Ca(NO3)2+2AgCl
M(CaCl2)=40+35.5*2=111 г/моль
M(AgNO3)=108+14+16*3=170 г/моль
n=M/m
n(CaCl2)=11.1/111=0.1 моль
n(AgNO3)=1.7/170=0.01 моль
M(AgCl)=108+35.5=143.5 г/моль
Далее считаем массу по недостатку
m(AgCl)=0.1*143.5=14.35 г

Answer 2

Calculating the Mass of Precipitate

To find the mass of the precipitate formed when a solution containing 11.1g of calcium chloride is mixed with a solution containing 1.7g of silver nitrate, we can use the concept of chemical reactions and precipitation. When calcium chloride and silver nitrate are mixed, they react to form calcium nitrate and silver chloride. The mass of the precipitate (silver chloride) can be calculated using stoichiometry.

The balanced chemical equation for the reaction is: CaCl2 + 2AgNO3 → Ca(NO3)2 + 2AgCl

From the equation, we can see that 1 mole of calcium chloride reacts with 2 moles of silver nitrate to produce 2 moles of silver chloride.

Using the given masses of the reactants, we can calculate the limiting reactant and then the mass of the precipitate formed.

First, we need to determine the limiting reactant by comparing the moles of each reactant.

The molar masses of the compounds are: - CaCl2: 40.08 + 35.45 = 75.53 g/mol - AgNO3: 107.87 + 14.01 + (3 * 16.00) = 169.87 g/mol

Now, let's calculate the moles of each reactant: - Moles of CaCl2 = 11.1g / 75.53 g/mol - Moles of AgNO3 = 1.7g / 169.87 g/mol

After determining the limiting reactant, we can calculate the mass of the precipitate formed.

Calculation Steps

1. Calculate the moles of each reactant. 2. Identify the limiting reactant. 3. Use the stoichiometry of the reaction to calculate the moles of AgCl formed. 4. Convert moles of AgCl to grams.

Let's proceed with the calculations.