At STP, which gas sample has a volume of 11.2 liters?
A
0.250 mole of NH3

B
0.500 mole of CO2

C
0.750 mole of NH3

D
1.00 mole of CO2

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

V = 0.25 0.0816 273 L

V = 5.56 L

B. 0.50 mole of

V = 0.50 0.0816 273 L

V = 11.13 L

C. 0.75 mole of

V = 0.75 0.0816 273 L

V = 16.70 L

D. 1 mole of

V = 1 0.0816 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:

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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)