Question

Помогите пожалуйста! Даю 40 баллов. №4. Рассчитайте объём сероводорода, который расходуется на взаимодействие с раствором с массовой долей гидроксида натрия 6% (плотность раствора 1,044 г /мл) объёмом 40 мл с образованием гидросульфита натрия. №5. Определите объём водорода (н.у.), необходимый для получения 1,8 • 1024 молекул воды.

Answer 1

Для начала посчитаем массу раствора NaOH

m(NaOH) = 40*1,044 = 41,76г

Содержание NaOH в растворе равно:

41,76/100*6 = 2,5г

Количество вещества NaOH равно:

2,5/40 = 0,0625моль

Уравнение реакции:

NaOH + H₂S → NaHS + H₂O

Следовательно сероводорода было также 0,0625моль, значит объем сероводорода равен:

0,0625*22,4 = 1,4л

2H₂ + O₂ → 2H₂O

Необходимо образоват 1,8*10²⁴ молекул воды, значит количество вещества воды равно: 1,8*10²⁴/6*10²³ = 3моль, судя по уравнению необходимо также 3 моль водорода. Значит объем водорода равен:

3*22,4 = 67,2л

Answer 2

Answer:

Question 4: To calculate the volume of hydrogen sulfide (H2S) consumed in the reaction with a solution of sodium hydroxide (NaOH) with a mass fraction of 6% (solution density of 1.044 g/ml) and a volume of 40 ml, we need to determine the number of moles of NaOH and then use the stoichiometry of the reaction to find the number of moles of H2S consumed.

First, let's calculate the number of moles of NaOH in the solution: - Mass of NaOH = mass fraction * mass of solution - Mass of solution = volume * density - Mass of NaOH = 0.06 * (40 ml * 1.044 g/ml)

Next, we need to convert the mass of NaOH to moles using its molar mass: - Molar mass of NaOH = 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) - Moles of NaOH = mass of NaOH / molar mass of NaOH

Finally, we can use the stoichiometry of the reaction to determine the moles of H2S consumed: - The balanced equation for the reaction is: 2 NaOH + H2S -> Na2S + 2 H2O - From the balanced equation, we can see that 2 moles of NaOH react with 1 mole of H2S. - Moles of H2S consumed = (moles of NaOH) / 2

Now we can calculate the volume of H2S consumed using its molar volume at standard temperature and pressure (STP), which is 22.4 L/mol: - Volume of H2S consumed = (moles of H2S consumed) * (22.4 L/mol)

Let's calculate the volume of H2S consumed using the given values:

1. Calculate the mass of NaOH: - Mass of NaOH = 0.06 * (40 ml * 1.044 g/ml) = 2.4944 g

2. Calculate the moles of NaOH: - Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol - Moles of NaOH = 2.4944 g / 40.00 g/mol = 0.06236 mol

3. Calculate the moles of H2S consumed: - Moles of H2S consumed = 0.06236 mol / 2 = 0.03118 mol

4. Calculate the volume of H2S consumed at STP: - Volume of H2S consumed = 0.03118 mol * 22.4 L/mol = 0.698 L

Therefore, the volume of hydrogen sulfide consumed in the reaction is approximately 0.698 liters.

Question 5: To determine the volume of hydrogen gas (H2) at standard temperature and pressure (STP) required to obtain 1.8 * 10^24 water (H2O) molecules, we need to use the Avogadro's law and the molar volume of a gas at STP.

Avogadro's law states that equal volumes of gases at the same temperature and pressure contain an equal number of molecules. Therefore, we can use the molar volume of a gas at STP to calculate the volume of hydrogen gas.

The molar volume of a gas at STP is 22.4 liters/mol. This means that 1 mole of any gas occupies 22.4 liters at STP.

To find the volume of hydrogen gas required, we need to determine the number of moles of hydrogen gas: - Moles of H2 = number of molecules / Avogadro's number

Once we have the moles of hydrogen gas, we can calculate the volume using the molar volume at STP: - Volume of H2 = moles of H2 * molar volume at STP

Let's calculate the volume of hydrogen gas required using the given values:

1. Calculate the number of moles of hydrogen gas: - Moles of H2 = 1.8 * 10^24 molecules / 6.022 * 10^23 molecules/mol = 2.99 mol

2. Calculate the volume of hydrogen gas at STP: - Volume of H2 = 2.99 mol * 22.4 L/mol = 67.1 L

Therefore, the volume of hydrogen gas (at STP) required to obtain 1.8 * 10^24 water molecules is approximately 67.1 liters.