Question

A reaction produced 37.5 L of oxygen gas at 307 K and 1.25 atm. How many moles of oxygen were produced? 0.538 mol, O2 1.86 mol, O2 3.22 mol, O2 7.56 mol O2

Answer 1

O 2 has a mass of 1.78g

Step-by-step explanation:

We are at STP and that means we have to use the ideal gas law equation!

P represents pressure (could have units of atm, depending on the units of the universal gas constant)

V represents volume (must have units of liters)

n represents the number of moles

R is the universal gas constant (has units of

L

× a t m

m o l

×

K

)

T represents the temperature, which must be in Kelvins.

Next, list your known and unknown variables. Our only unknown is the number of moles of

O

2

(

g

)

. Our known variables are P,V,R, and T.

At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821

L

×

a

t

m

m

o

l

×

K

Now we have to rearrange the equation to solve for n

n

=

P

V

R

T

n

=

1

atm

×

1.25

L

0.0821

Lxxatm

m

o

l

×

K

×

273

K

n

=

0.05577

m

o

l

To get the mass of

O

2

, we just have to use the molar mass of oxygen as a conversion factor:

0.0577

mol

O

2

×

32.00

g

1

mol

= 1.78g

O

2

Answer 2

Answer : The correct option is, 1.86 moles

Solution :

using ideal gas equation,

`PV=nRT`

where,

n = number of moles of gas = ?

P = pressure of the gas = 1.25 atm

T = temperature of the gas = 307 K

R = gas constant = 0.0821 Latm/moleK

V = volume of gas = 37.5 L

Now put all the given values in the above equation, we get the moles of gas.

`1.25atm* 37.5L=n* 0.0821L.atm/moleK* 307K`

`n=1.86mole`

Therefore, the moles of oxygen produced were, 1.86 moles