Question

Определите какой объём при нормальных условиях занимает 46,4 г бутана C4H10

Answer 1

Дано:
m(C4H10)=46,4 г. n(C4H10) = m(C4H10) / M(C4H10) = 46,4 / 58=0,8 моль
М(С4Н10)=58 г
Vm=22,4 л/моль. V(C4H10) = Vm* n(C4H10)= 22,4 * 0,8 = 17,92 л
Найти:
V(C4H10)=?

Answer 2

Volume of 46.4g of Butane at Standard Conditions

The volume of 46.4g of butane (C4H10) at standard conditions can be calculated using the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Using the molar mass of butane (C4H10), which is approximately 58.12 g/mol, we can calculate the number of moles of butane present in 46.4g.

Calculating the Number of Moles: The number of moles (n) of butane can be calculated using the formula: n = mass / molar mass n = 46.4g / 58.12 g/mol n ≈ 0.799 moles

Now, using the ideal gas law, we can calculate the volume of butane at standard conditions.

Applying the Ideal Gas Law: The ideal gas law equation can be rearranged to solve for volume: V = (nRT) / P

Where: - n = 0.799 moles (calculated above) - R = 0.0821 L·atm/(K·mol) (gas constant) - T = 273 K (standard temperature) - P = 1 atm (standard pressure)

Calculating the Volume: V = (0.799 moles * 0.0821 L·atm/(K·mol) * 273 K) / 1 atm V ≈ 18.6 liters

Therefore, the volume of 46.4g of butane at standard conditions (1 atm and 273 K) is approximately 18.6 liters.