If one mol of gas = 22.4 at STP how many liters is occupied by 3.65 moles of oxygen gas at STP

1 mole gas at STP = 22.4 L

so volume of oxygen at STP:

1)

moles of oxygen (n) = 1.86 moles

Explanation:

according to  Boyle's law the formula to solve this problem is:

PV=nRt

when P is the pressure which equal 1.25 atm

and V is the volume which equal 37.5 L

n is the number of moles which we need to calculate it

R is constant which equal 0.082

t is the temperature in kelvin

By substitution:

1.25*37.5 = n * 0.082 * 307

So n = 1.86 moles

2)

the volume of oxygen gas = 34 L

Explanation:

at standard temperature and pressure (STP) 1 mole of gas will equal = 22.4L

So when we have 1.5 moles of oxygen at standard temperature and pressure (STP) so we will estimate it like that

1.5 moles *22.4 L/ 1 mole = approximately 34 L

3)

The volume of H2 = 2.29 L

Explanation:

according to the Balanced equation we can see that the molar ratio between Zn and H2 1 : 1

so to know the number of moles of H2 we will get it for Zn first :

number of moles Zn = mass of Zn / molar mass Zn

                             = 5.98 / 65.39 =0.0914 moles

so number of moles H2 = 0.09 moles

by  substitution in the following formula:

PV = nRT

0.978 * V = 0.09 * 0.082 * 298

so The volume of H2 = 2.29 L

4)

Volume of O2 = 1.4 L

Explanation:

first we have to balance  the equation:

2Na2O2 +2CO2 → 2Na2CO3 + O2

2 mole CO2 give 1 mole of O2 so the molar ratio is 2:1

at STP 1 mole of gas will equal = 22.4 L

             ??? moles of CO2 = 2.8

n CO2 = 0.125 moles so n O2 = 0.125 /2 = 0.0625 moles

so when 1 mole of as = 22.4

              0.0625 moles O2 = ???

Volume of O2 =0.0625 moles * 22.4 L/ 1 mole

                        = 1.4 L

5)

the initial quantity of sodium metal used = 17.2 gram

Explanation:

at STP 1 mole of gas will equal = 22.4 L

so  moles of H2 equal ?? when 8.40 liters of H2 gas were produced

so moles of H2  = 8.4/22.4 =0.375 moles

and according to the balanced equation the molar ratio between H2 ans Na is 1 : 2

so number of moles for Na = 0.375 *2 = 0.75 moles

to get the initial quantity of sodium metal (mass Na) = number of moles * molar mass

mass Na = 0.75 moles * 23 gm/mole= 17.25

6)

False

Explanation:

because STP means standard temperature and pressure. and the Standard temperature must be 273 K and the standard pressure must be 1 atm but in the question the temperature is 298 K not 273 K so , It is not a standard temperature

7)

22.4 liters

Explanation:

1 mole of gas will equal = 22.4 L

because at the Standard temperature must be 273 K and the standard pressure must be 1 atm

so V = nRT/P

       =1 mole * 0.082 * 273 K / 1 aTm

        = 22.4 liters

1)

moles of oxygen (n) = 1.86 moles

Explanation:

according to  Boyle's law the formula to solve this problem is:

PV=nRt

when P is the pressure which equal 1.25 atm

and V is the volume which equal 37.5 L

n is the number of moles which we need to calculate it

R is constant which equal 0.082

t is the temperature in kelvin

By substitution:

1.25*37.5 = n * 0.082 * 307

So n = 1.86 moles

2)

the volume of oxygen gas = 34 L

Explanation:

at standard temperature and pressure (STP) 1 mole of gas will equal = 22.4L

So when we have 1.5 moles of oxygen at standard temperature and pressure (STP) so we will estimate it like that

1.5 moles *22.4 L/ 1 mole = approximately 34 L

3)

The volume of H2 = 2.29 L

Explanation:

according to the Balanced equation we can see that the molar ratio between Zn and H2 1 : 1

so to know the number of moles of H2 we will get it for Zn first :

number of moles Zn = mass of Zn / molar mass Zn

                             = 5.98 / 65.39 =0.0914 moles

so number of moles H2 = 0.09 moles

by  substitution in the following formula:

PV = nRT

0.978 * V = 0.09 * 0.082 * 298

so The volume of H2 = 2.29 L

4)

Volume of O2 = 1.4 L

Explanation:

first we have to balance  the equation:

2Na2O2 +2CO2 → 2Na2CO3 + O2

2 mole CO2 give 1 mole of O2 so the molar ratio is 2:1

at STP 1 mole of gas will equal = 22.4 L

             ??? moles of CO2 = 2.8

n CO2 = 0.125 moles so n O2 = 0.125 /2 = 0.0625 moles

so when 1 mole of as = 22.4

              0.0625 moles O2 = ???

Volume of O2 =0.0625 moles * 22.4 L/ 1 mole

                        = 1.4 L

5)

the initial quantity of sodium metal used = 17.2 gram

Explanation:

at STP 1 mole of gas will equal = 22.4 L

so  moles of H2 equal ?? when 8.40 liters of H2 gas were produced

so moles of H2  = 8.4/22.4 =0.375 moles

and according to the balanced equation the molar ratio between H2 ans Na is 1 : 2

so number of moles for Na = 0.375 *2 = 0.75 moles

to get the initial quantity of sodium metal (mass Na) = number of moles * molar mass

mass Na = 0.75 moles * 23 gm/mole= 17.25

6)

False

Explanation:

because STP means standard temperature and pressure. and the Standard temperature must be 273 K and the standard pressure must be 1 atm but in the question the temperature is 298 K not 273 K so , It is not a standard temperature

7)

22.4 liters

Explanation:

1 mole of gas will equal = 22.4 L

because at the Standard temperature must be 273 K and the standard pressure must be 1 atm

so V = nRT/P

       =1 mole * 0.082 * 273 K / 1 aTm

        = 22.4 liters

44.8 L of N₂.

Explanation:

We'll begin by writing the balanced equation for the. This is illustrated below:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

1 mole of N₂ reacted to produce 2 moles of NH₃.

Therefore, Xmol of N₂ will react to produce 4 moles of NH₃ i.e

Xmol of N₂ = (1 × 4)/2

Xmol of N₂ = 2 moles

Thus, 2 moles of N₂ reacted to produce 4 moles of NH₃.

Finally, we shall determine the volume of N₂ required for the reaction. This can be obtained as follow:

1 mole of N₂ occupies 22.4 L at STP.

Therefore, 2 moles of N₂ will occupy = (2 × 22.4) = 44.8 L

Thus, 44.8 L of N₂ is needed to produce 4 moles of NH₃.

44.8 L of N₂.

Explanation:

We'll begin by writing the balanced equation for the. This is illustrated below:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

1 mole of N₂ reacted to produce 2 moles of NH₃.

Therefore, Xmol of N₂ will react to produce 4 moles of NH₃ i.e

Xmol of N₂ = (1 × 4)/2

Xmol of N₂ = 2 moles

Thus, 2 moles of N₂ reacted to produce 4 moles of NH₃.

Finally, we shall determine the volume of N₂ required for the reaction. This can be obtained as follow:

1 mole of N₂ occupies 22.4 L at STP.

Therefore, 2 moles of N₂ will occupy = (2 × 22.4) = 44.8 L

Thus, 44.8 L of N₂ is needed to produce 4 moles of NH₃.

The correct answer is C. 44.8 L

Explanation:

22.4 × 2 = 44.8

The correct answer is C. 44.8 L

Explanation:

22.4 × 2 = 44.8

Answer:

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm