How many moles of gas are required to exert 15.4 atm of pressure in a 2.92 L container at 32.0°C

Chemistry

1 answer:

Iteru [2.4K]3 years ago

8 0

Answer:

1.8 moles

Explanation:

Number of moles = ?

Pressure = 15.4 atm

Volume = 2.92L

Temperature = 32°C = (32 + 273.15)K = 305.15K

From ideal gas equation,

PV = nRT

P = pressure of the ideal gas

V = volume in which the gas occupies

n = number of moles

R = ideal gas constant = 0.082J/mol.K

T = temperature of the gas

PV = nRT

n = PV / RT

n = (15.4 * 2.92) / (0.082 * 305.15)

n = (44.968 / 25.0223)

n = 1.797 moles = 1.8 moles

Number of moles of the gas is 1.8 moles

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A gas mixture contains 3.00 atm of H2 and 1.00 atm of O2 in a 1.00 L vessel at 400K. If the mixture burns to form water while th

sleet_krkn [62]

Answer:

$p_{H_2O}=2.00atm$

Explanation:

Hello!

In this case, according to the following chemical reaction:

$2H_2+O_2\rightarrow 2H_2O$

It means that we need to compute the moles of hydrogen and oxygen that are reacting, via the ideal gas equation as we know the volume, pressure and temperature:

$n_{H_2}=\frac{3.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0914molH_2 \\\\n_{O_2}=\frac{1.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0305molH_2$

Thus, the yielded moles of water are computed by firstly identifying the limiting reactant:

$n_{H_2O}^{by\ H_2} = 0.0914molH_2*\frac{2molH_2O}{2molH_2} =0.0914molH_2O\\\\n_{H_2O}^{by\ O_2} = 0.0305molO_2*\frac{2molH_2O}{1molO_2} =0.0609molH_2O$

Thus, the fewest moles of water are 0.0609 mol so the limiting reactant is oxygen; in such a way, by using the ideal gas equation once again, we compute the pressure of water:

$p_{H_2O}=\frac{0.0609molH_2O*0.08206\frac{atm*L}{mol*K}*400K}{1.00L}\\\\ p_{H_2O}=2.00atm$