Question

Вычислите V(объем) H2, который получается в результате дегидрирования этана массой 3 грамма ? дайте пожалуйста подробный ответ

Answer 1

Дано
m(C2H6)=3 г
V(H2)-?
Решение:
C2H6=С2H4+H2
n(C2H6)=m/M=3/2*12+6*1=3/30=0,1 моль
По уравнению реакции
n(C2H6)=n(H2)=0,1 моль
V(H2)=n*Vm=0,1*22,4=2,24 л
Ответ: объем водорода равен 2,24 л

Answer 2

Calculating the Volume of H2 from the Dehydration of Ethane

To calculate the volume of hydrogen gas (H2) produced from the dehydration of 3 grams of ethane, we can use the stoichiometry of the reaction. The balanced chemical equation for the dehydration of ethane is:

C2H6 → C2H4 + H2

From this equation, we can see that 1 mole of ethane (C2H6) produces 1 mole of hydrogen gas (H2).

Calculation Steps

1. Calculate the moles of ethane (C2H6) using its molar mass. 2. Use the mole ratio from the balanced chemical equation to find the moles of hydrogen gas (H2). 3. Convert the moles of hydrogen gas to volume using the ideal gas law.

Step 1: Calculate the Moles of Ethane (C2H6)

The molar mass of ethane (C2H6) is 30.07 g/mol.

Moles of C2H6 = Mass of C2H6 / Molar mass of C2H6 Moles of C2H6 = 3 g / 30.07 g/mol Moles of C2H6 = 0.0998 mol

Step 2: Use the Mole Ratio to Find the Moles of Hydrogen Gas (H2)

From the balanced chemical equation, the mole ratio of C2H6 to H2 is 1:1.

Moles of H2 = Moles of C2H6 Moles of H2 = 0.0998 mol

Step 3: Convert Moles of Hydrogen Gas to Volume

We can use the ideal gas law to convert moles of hydrogen gas to volume at standard temperature and pressure (STP), which is 0°C and 1 atm.

The ideal gas law is given by: PV = nRT Where: P = Pressure (1 atm) V = Volume n = Moles of gas R = Ideal gas constant (0.0821 L·atm/mol·K) T = Temperature in Kelvin (273 K at STP)

Using the ideal gas law, we can solve for the volume (V) of hydrogen gas (H2).

V = (nRT) / P V = (0.0998 mol * 0.0821 L·atm/mol·K * 273 K) / 1 atm V ≈ 2.19 L

The volume of hydrogen gas (H2) produced from the dehydration of 3 grams of ethane is approximately 2.19 liters.