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:

link

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)

the answer

explanation:

the capillary tube was too close to the bottom of the beaker.

the ruler may have moved.

water got into the capillary tube.

the temperature was not allowed to equilibrate in the 2-4 minutes.