Calculate Oxygen Mass at RT & Pressure: 78 cm3

  • Context: MHB 
  • Thread starter Thread starter markosheehan
  • Start date Start date
  • Tags Tags
    Mass Oxygen Pressure
Click For Summary
SUMMARY

The discussion focuses on calculating the mass of 78 cm³ of oxygen gas (O₂) at room temperature and pressure. The correct method involves determining the number of moles using the molar gas volume, which is 24 liters at standard conditions. The calculation yields 0.104 grams when applying the formula $m(O) = n(O)M(O)$, where M(O) is the molar mass of oxygen (32 g/mol). Participants clarify that the reaction involving hydrogen peroxide (H₂O₂) produces half a mole of O₂, but the mass calculation remains based on the full molar mass of O₂.

PREREQUISITES
  • Understanding of gas laws and molar volume
  • Familiarity with stoichiometry in chemical reactions
  • Knowledge of the molar mass of oxygen (32 g/mol)
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the Ideal Gas Law and its applications in calculations
  • Learn about stoichiometric calculations in chemical reactions
  • Explore the concept of molar volume for different gases
  • Investigate the decomposition of hydrogen peroxide and its products
USEFUL FOR

Chemistry students, educators, and professionals involved in chemical calculations, particularly those focusing on gas laws and stoichiometry.

markosheehan
Messages
133
Reaction score
0
Calculate the mass of 78 cm3 of oxygen at room temperature and pressure .

Im trying to work out the number of moles and then multiply by 32.
 
Mathematics news on Phys.org
markosheehan said:
Calculate the mass of 78 cm3 of oxygen at room temperature and pressure .

Im trying to work out the number of moles and then multiply by 32.

$n(O) = \dfrac{V(O)}{V_m(O)}$

$m(O) = n(O)M(O)$.

$V_m$ is the molar gas volume.
 
Im sorry I am still confused. I know the right answer is .104g
 
markosheehan said:
Im sorry I am still confused. I know the right answer is .104g

The molar volume of a gas ($V_m$) at room temperature and standard pressure is $24$ liter.
That is, $1$ mole is $24$ liter.
How many moles does that make in $78\text{ cm}^3$?
 
Thanks.
I feel the answer is wrong. 78/24000 ×32 is what gives the answer at the back of the book. Should it not be multiplied by 16 though because in the balanced equation for the reaction it's a half mile of oxygen is formed.?
 
markosheehan said:
Thanks.
I feel the answer is wrong. 78/24000 ×32 is what gives the answer at the back of the book. Should it not be multiplied by 16 though because in the balanced equation for the reaction it's a half mile of oxygen is formed.?

Not sure what you mean. Was there a reaction involved?

We have 78/24000 moles of oxygen molecules ($O_2$).
Each molecule consists of 2 oxygen atoms, and an oxygen atom has mass 16u.
Therefore we multiply the moles with $2\times 16=32$.
 
sorry i forgot to mention it was calculate the mass of 78 cm3 of oxygen produced by the the composition of hydrogen peroxide.
h202 goes to h20 +.5o2
so does this not mean you would multiply it by 16 instead of 32
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K