Question

Напишите уравнения возможных реакций между следующими веществами:А) водород и оксид железа В) водород и хлорБ) соляная кислота и алюминий Г) водород и оксид серебра4. Решите схему превращений: Н2->HCl->ZnCl2->AgNO3->NaNO35.Решить задачу:Смешали нитрат серебра массой 15 г с иодидом натрия массой 4,25 г. Определите массу осадка. ПОМОГИТЕ!!!!!5 ЗАДАНИЕ МНЕ ОБЯЗАТЕЛЬНО НАДО СДАТЬ, ПОЖАЛУЙСТА!!!!!!

Answer 1

А)

FeO + H2 = Fe + H2O

Fe2O3 + 3H2 = 2Fe + 3H2O

Б)

Al + 6HCl = 2AlCl3 + 3H2

В)

H2 + Cl2 = 2HCl

Г)

Ag2O + H2 = 2Ag + H2O

4.

H2 + Cl2 = 2HCl

ZnO + 2HCl = ZnCl2 + H2O

ZnCl2 -> AgNO3 - нет решения

AgNO3 + NaCl = AgCl + NaNO3

5.

Дано

m(AgNO3) = 15 г

m(NaI) = 4.25 г

Решение

n(AgNO3) = m(AgNO3) / M(AgNO3) = 15 / (108 + 14 + 3 * 16) = 15 / 170 = 0.09 моль

n(NaI) = m(NaI) / M(NaI) = 4.25 / (23 + 127) = 4.25 / 150 = 0.03 моль

AgNO3 + NaI = AgI + NaNO3

NaI - лимитирующий реагент, так как на 0.09 моль AgNO3 требуется 0.09 моль NaI, а дано только 0.03 моль, поэтому:

n(AgI) = n(NaI) = 0.03 моль

m(AgI) = n(AgI)M(AgI) = 0.03 * (108 + 127) = 0.03 * 235 = 7.05 г

Ответ: 7.05 г

Answer 2

Equations for Possible Reactions:

A) Hydrogen and Iron Oxide: - The possible reaction between hydrogen and iron oxide is the reduction of iron oxide to iron and the formation of water. The balanced equation for this reaction is: Fe2O3 + 3H2 → 2Fe + 3H2O [[1]]

B) Hydrogen and Chlorine: - The possible reaction between hydrogen and chlorine is the formation of hydrogen chloride. The balanced equation for this reaction is: H2 + Cl2 → 2HCl [[2]]

C) Hydrochloric Acid and Aluminum: - The possible reaction between hydrochloric acid and aluminum is the displacement of hydrogen from hydrochloric acid by aluminum. The balanced equation for this reaction is: 2Al + 6HCl → 2AlCl3 + 3H2 [[3]]

D) Hydrogen and Silver Oxide: - The possible reaction between hydrogen and silver oxide is the reduction of silver oxide to silver and the formation of water. The balanced equation for this reaction is: 2Ag2O + H2 → 4Ag + H2O [[4]]

Scheme of Transformations:

The given scheme of transformations is as follows: - H2 -> HCl -> ZnCl2 -> AgNO3 -> NaNO3

To solve this scheme, we need to determine the products of each reaction.

1. H2 -> HCl: - The reaction between hydrogen and chlorine produces hydrogen chloride. H2 + Cl2 → 2HCl

2. HCl -> ZnCl2: - The reaction between hydrochloric acid and zinc produces zinc chloride and hydrogen gas. Zn + 2HCl → ZnCl2 + H2

3. ZnCl2 -> AgNO3: - The reaction between zinc chloride and silver nitrate produces silver chloride and zinc nitrate. ZnCl2 + 2AgNO3 → 2AgCl + Zn(NO3)2

4. AgNO3 -> NaNO3: - The reaction between silver nitrate and sodium chloride produces silver chloride and sodium nitrate. AgNO3 + NaCl → AgCl + NaNO3

Therefore, the final product of the given scheme is sodium nitrate (NaNO3).

Problem Solution:

We are given the following masses: - Mass of silver nitrate (AgNO3) = 15 g - Mass of sodium iodide (NaI) = 4.25 g

To determine the mass of the precipitate formed, we need to identify the reaction between silver nitrate and sodium iodide and calculate the mass of the precipitate formed.

The balanced equation for the reaction between silver nitrate and sodium iodide is: AgNO3 + NaI → AgI + NaNO3

From the balanced equation, we can see that silver iodide (AgI) is the precipitate formed.

To calculate the mass of the precipitate, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

To find the limiting reactant, we can compare the number of moles of each reactant using their molar masses.

The molar mass of AgNO3 = 169.87 g/mol The molar mass of NaI = 149.89 g/mol

Using the given masses, we can calculate the number of moles of each reactant: - Moles of AgNO3 = Mass of AgNO3 / Molar mass of AgNO3 - Moles of NaI = Mass of NaI / Molar mass of NaI

After calculating the moles of each reactant, we can determine the limiting reactant by comparing the mole ratios from the balanced equation.

The balanced equation shows that 1 mole of AgNO3 reacts with 1 mole of NaI to produce 1 mole of AgI.

Therefore, the limiting reactant is the reactant that produces the least amount of AgI.

Once we determine the limiting reactant, we can calculate the mass of the precipitate formed using the mole ratio from the balanced equation.

Finally, we can calculate the mass of the precipitate using the molar mass of AgI.

Please provide the molar masses of AgNO3 and NaI to proceed with the calculations.