Question

Химия,50 баллов,задача. Сколько грамм AgNO3 выпадет в осадок при взаимодействии 50 грамм оксида серебра с азотной кислотой.

Answer 1

Нитрат серебра растворимый

Answer 2

1) Вычисляем массы 1 моля интересующих нас веществ:

М(Ag₂O) = 2*108 + 16 = 232 г/моль

М(AgNO₃) = 108 + 14 + 3*16 = 170 г/моль

2) Уравнение реакции:

Ag₂O + 2 HNO₃ → 2 AgNO₃ ↓ + H₂O

По уравнению видно, что из 1 моля Ag₂O (или 232 г) образуется 2 моля AgNO₃ (2*170 = 340 г)

Поэтому можно составить пропорцию:

Из 232 г Ag₂O образуется 340 г AgNO₃

Из 50 г Ag₂O  -------------------   Х г AgNO₃

Х = 50*340/232 = 73,28 г

Ответ: 73,28 г AgNO₃

Answer 3

Problem Analysis

To find out how many grams of AgNO3 will precipitate when 50 grams of silver oxide (Ag2O) reacts with nitric acid (HNO3), we need to determine the stoichiometry of the reaction. The balanced chemical equation for the reaction between silver oxide and nitric acid is:

2 Ag2O + 4 HNO3 -> 4 AgNO3 + H2O

According to this equation, 2 moles of Ag2O react with 4 moles of HNO3 to produce 4 moles of AgNO3. We can use this stoichiometry to calculate the amount of AgNO3 that will precipitate.

Calculation

To calculate the amount of AgNO3 that will precipitate, we need to follow these steps:

1. Convert the mass of Ag2O to moles using its molar mass. 2. Use the stoichiometry of the reaction to determine the moles of AgNO3 produced. 3. Convert the moles of AgNO3 to grams using its molar mass.

Let's perform the calculations step by step.

Step 1: Convert the mass of Ag2O to moles The molar mass of Ag2O is the sum of the atomic masses of silver (Ag) and oxygen (O): Ag: 107.87 g/mol O: 16.00 g/mol

Molar mass of Ag2O = (2 * Ag) + O = (2 * 107.87) + 16.00 = 231.74 g/mol

To convert the mass of Ag2O to moles, we divide the mass by the molar mass: Moles of Ag2O = Mass of Ag2O / Molar mass of Ag2O = 50 g / 231.74 g/mol

Step 2: Use the stoichiometry of the reaction to determine the moles of AgNO3 produced According to the balanced equation, 2 moles of Ag2O react to produce 4 moles of AgNO3. Therefore, the moles of AgNO3 produced can be calculated as follows: Moles of AgNO3 = (Moles of Ag2O / 2) * 4

Step 3: Convert the moles of AgNO3 to grams The molar mass of AgNO3 is the sum of the atomic masses of silver (Ag), nitrogen (N), and oxygen (O): Ag: 107.87 g/mol N: 14.01 g/mol O: 16.00 g/mol

Molar mass of AgNO3 = Ag + (N * 2) + (O * 3) = 107.87 + (14.01 * 2) + (16.00 * 3) = 169.87 g/mol

To convert the moles of AgNO3 to grams, we multiply the moles by the molar mass: Mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

Now, let's perform the calculations.

Calculation Results

Step 1: Convert the mass of Ag2O to moles Molar mass of Ag2O = 231.74 g/mol Moles of Ag2O = 50 g / 231.74 g/mol = 0.215 moles

Step 2: Use the stoichiometry of the reaction to determine the moles of AgNO3 produced Moles of AgNO3 = (0.215 moles / 2) * 4 = 0.430 moles

Step 3: Convert the moles of AgNO3 to grams Molar mass of AgNO3 = 169.87 g/mol Mass of AgNO3 = 0.430 moles * 169.87 g/mol = 73.06 grams

Answer

When 50 grams of silver oxide (Ag2O) reacts with nitric acid (HNO3), approximately 73.06 grams of AgNO3 will precipitate as a result of the reaction.

Please note that the calculations provided are based on the given balanced chemical equation and the assumption that the reaction goes to completion.