Respuesta :
Answer:
0.500 moles of CO2 has a volume of 11.2 L at STP (option B)
Explanation:
Step 1: Data given
Volume of a gas at STP = 11.2 L
STP: Pressure = 1 atm and temperature = 273 K
Step 2: Calculate volume
p*V= n*R*T
V = (n*R*T)/p
⇒with V = the volume of the gas = TO BE DETERMINED
⇒with n = the number of moles of the gas
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 273 K
⇒with p = the pressure of the gas = 1 atm
A ) 0.250 mole of NH3
V = (0.250 * 0.08206 * 273) / 1
V = 5.6 L
B ) 0.500 mole of CO2
V = (0.500 * 0.08206 * 273) / 1
V = 11.2 L
C ) 0.750 mole of NH3
V = (0.750 * 0.08206 * 273) / 1
V = 16.8 L
D) 1.00 mole of CO2
V = (1.00 * 0.08206* 273) / 1
V = 22.4 L
0.500 moles of CO2 has a volume of 11.2 L at STP (option B)
From the given gases, 0.5 moles of Carbon dioxide has a volume of 11.2 L. Thus, option B is correct.
The volume of gas at STP can be calculated with the help of the ideal gas equation. According to the equation:
PV = nRT
where, P= pressure = 1 atm.
V = volume (L)
n = moles of the gas
R = constant = 0.0816 L atm/mol K
T = temperature = 273 K
The volume of the given gases will be:
A. 0.25 mole of [tex]\rm NH_3[/tex]
V = 0.25 [tex]\times[/tex] 0.0816 [tex]\times[/tex] 273 L
V = 5.56 L
B. 0.50 mole of [tex]\rm CO_2[/tex]
V = 0.50 [tex]\times[/tex] 0.0816 [tex]\times[/tex] 273 L
V = 11.13 L
C. 0.75 mole of [tex]\rm NH_3[/tex]
V = 0.75 [tex]\times[/tex] 0.0816 [tex]\times[/tex] 273 L
V = 16.70 L
D. 1 mole of [tex]\rm CO_2[/tex]
V = 1 [tex]\times[/tex] 0.0816 [tex]\times[/tex] 273 L
V = 22.27 L
Thus, from the given gases, 0.5 moles of Carbon dioxide has volume of 11.2 L. Thus, option B is correct.
For more information about volume at STP, refer to the link:
https://brainly.com/question/24050436
