Question

Определите объём метана (н.у),который можно получить из 36 г карбида алюминия с массовой долей примесей 20 %.

Answer 1

Решение в следующей картинке

Answer 2

Calculating the Volume of Methane (CH4) Produced from Aluminum Carbide (Al4C3)

To calculate the volume of methane (CH4) produced from 36g of aluminum carbide (Al4C3) with a 20% impurity, we can use the following chemical equation:

Al4C3 + 12H2O → 4Al(OH)3 + 3CH4

First, we need to calculate the molar mass of aluminum carbide and the amount of pure aluminum carbide present in 36g with 20% impurity.

The molar mass of Al4C3 is approximately 144 g/mol.

The pure mass of aluminum carbide present in 36g with 20% impurity can be calculated as follows:

36g * (1 - 20%) = 36g * 0.8 = 28.8g

Now, we can calculate the moles of aluminum carbide:

moles = mass / molar mass moles = 28.8g / 144 g/mol moles ≈ 0.2 moles

According to the balanced chemical equation, 1 mole of Al4C3 produces 3 moles of CH4.

So, the moles of CH4 produced from 0.2 moles of Al4C3 can be calculated as:

moles of CH4 = 0.2 moles * 3 moles of CH4 = 0.6 moles

Now, we can use the ideal gas law to calculate the volume of CH4 produced at standard temperature and pressure (STP), which is 0°C and 1 atm.

The molar volume of a gas at STP is approximately 22.4 L/mol.

Therefore, the volume of CH4 produced can be calculated as:

volume = moles * molar volume volume = 0.6 moles * 22.4 L/mol volume ≈ 13.44 L

So, the volume of methane (CH4) that can be obtained from 36g of aluminum carbide with a 20% impurity is approximately 13.44 liters.