Respuesta :

Answer:

The volume is occupies by 21,0g of methane is 32, 3 L.

Explanation:

We use the formula PV=nRT. We convert the unit temperature Celsiud into Kelvin: 0°C=273 K---> 27°C= 27+273= 300K. We calculate the weight of 1 mol of CH4:

Weight 1 mol CH4= Weight C+ (Weight H)x4= 12 g+ 1g x4= 16 g

16 g---1 mol CH4

21g---x= (21g x 1 mol CH4)/16g= 1, 31 mol CH4

PV=nRT    ----> V= (nRT)/P

V= (1, 31 mol x 0,082 l atm/K mol x 300 K)/ 1 atm

V= 32, 3875 L

samueladesida43

The volume occupied by 21.0 g of methane (CH4) at 27 degree celsius and 1 atm is 32.3L

IDEAL GAS LAW:

  • The volume occupied by a gas can be calculated by using the ideal gas law equation as follows:

PV = nRT

Where;

  1. P = pressure (atm)
  2. V = volume (L)
  3. n = number of moles (mol)
  4. R = gas law constant (0.0821 Latm/molK)

  • According to this question;
  1. n = 21g ÷ 16g/mol = 1.3125mol
  2. V = ?
  3. P = 1 atm
  4. T = 27°C + 273 = 300K

  • 1 × V = 1.3125 × 0.0821 × 300

  • V = 32.32L

  • Therefore, volume occupied by 21.0 g of methane (CH4) at 27 degree celsius and 1 atm is 32.3L.

Learn more about ideal gas law at: https://brainly.com/question/4147359?referrer=searchResults

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