Question

Здравствуйте! Пожалуйста, помогите решить 2 задачи по химии (гимназия). 1) Для сжигания скольких молей пентана потребуется 40 моль кислорода? Сколько кубических метров воздуха содержат такое же количество кислорода? 2) Сколько кубических сантиметров гексана (р=0,660г/см3) можно сжечь с помощью 1064 л кислорода? Спасибо за помощь!!

Answer 1

Ответ:

1 моль пентана сжигается с 5 молями кислорода, так что 40 моль кислорода могут сжечь 8 молей пентана (40/5 = 8).

1 кубический метр воздуха содержит примерно 0,2% кислорода, так что 40 моль кислорода равны 200 кубическим метрам воздуха (40/0.002 = 20000)

Объяснение:

Перед началом расчета, необходимо перевести объем кислорода из литров в кубические сантиметры:

1064 л = 1064 * 1000 см3 = 1064000 см3

После этого, мы можем рассчитать количество гексана, который может быть сжеган:

1064000 см3 кислорода может сжечь 1064000 / 5 = 212800 см3 пентана

Объем гексана можно рассчитать, используя его плотность:

212800 см3 пентана = 212800 * 0.660 г/см3 = 140632 грамм

Таким образом, мы можем рассчитать количество кубических сантиметров гексана:

140632 грамм = 140632 / 0.660 г/см3 = 212318.18 см3

Answer 2

Task 1: Сжигание пентана

To determine the number of moles of pentane required to burn 40 moles of oxygen, we need to consider the balanced chemical equation for the combustion of pentane:

C5H12 + 8O2 -> 5CO2 + 6H2O

From the equation, we can see that 1 mole of pentane (C5H12) reacts with 8 moles of oxygen (O2). Therefore, to find the number of moles of pentane required, we can set up a proportion:

1 mole of pentane / 8 moles of oxygen = x moles of pentane / 40 moles of oxygen

Solving for x, we find:

x = (1 mole of pentane / 8 moles of oxygen) * 40 moles of oxygen x = 5 moles of pentane

Therefore, to burn 40 moles of oxygen, we would need 5 moles of pentane.

To calculate the volume of air containing the same amount of oxygen, we need to know the composition of air. Air is approximately 21% oxygen, 78% nitrogen, and 1% other gases. Therefore, we can calculate the volume of air using the following proportion:

21% oxygen in air / 100% oxygen = volume of air / volume of oxygen

Substituting the values, we find:

volume of air = (21% oxygen in air / 100% oxygen) * volume of oxygen

Let's calculate the volume of air containing 40 moles of oxygen:

volume of air = (21/100) * 40 moles of oxygen volume of air = 8.4 moles of oxygen

Therefore, the volume of air containing the same amount of oxygen as 40 moles of oxygen is 8.4 moles.

Please note that the volume of air is given in moles, not in cubic meters. If you need to convert the volume of air from moles to cubic meters, you would need to know the temperature and pressure conditions.

Task 2: Сжигание гексана

To determine the number of cubic centimeters of hexane that can be burned with 1064 liters of oxygen, we need to consider the balanced chemical equation for the combustion of hexane:

C6H14 + 19/2O2 -> 6CO2 + 7H2O

From the equation, we can see that 1 mole of hexane (C6H14) reacts with 19/2 moles of oxygen (O2). Therefore, to find the number of moles of hexane required, we can set up a proportion:

1 mole of hexane / (19/2) moles of oxygen = x moles of hexane / 1064 liters of oxygen

Solving for x, we find:

x = (1 mole of hexane / (19/2) moles of oxygen) * 1064 liters of oxygen x = 56 moles of hexane

To calculate the mass of hexane, we can use the density of hexane, which is given as 0.660 g/cm^3. The formula to calculate the mass is:

mass = density * volume

Substituting the values, we find:

mass = 0.660 g/cm^3 * 56 moles of hexane mass = 36.96 g of hexane

To convert the mass of hexane to cubic centimeters, we need to use the molar mass of hexane, which is approximately 86.18 g/mol. The formula to convert mass to volume is:

volume = mass / density

Substituting the values, we find:

volume = 36.96 g of hexane / 0.660 g/cm^3 volume = 56 cm^3 of hexane

Therefore, with 1064 liters of oxygen, we can burn approximately 56 cubic centimeters of hexane.

I hope this helps! Let me know if you have any further questions.