Calculate Oxygen Mass at RT & Pressure: 78 cm3

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Discussion Overview

The discussion revolves around calculating the mass of 78 cm3 of oxygen gas at room temperature and pressure. Participants are exploring the relationship between volume, moles, and mass, with references to molar volume and chemical reactions.

Discussion Character

  • Mathematical reasoning
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • Some participants propose using the formula $n(O) = \dfrac{V(O)}{V_m(O)}$ to calculate the number of moles of oxygen.
  • Others mention that the molar volume of a gas at room temperature and standard pressure is 24 liters, leading to questions about how many moles correspond to 78 cm3.
  • One participant asserts that the correct mass is 0.104g, referencing a calculation involving the molar mass of oxygen.
  • Another participant challenges the previous calculations, suggesting that the mass should be calculated using 16 instead of 32 due to the stoichiometry of the reaction involving hydrogen peroxide decomposition.
  • There is a mention of the balanced equation for the reaction, indicating that half a mole of oxygen is produced, which raises questions about the appropriate multiplier for the mass calculation.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating the mass of oxygen, with no consensus reached on whether to use 32 or 16 as the multiplier based on the reaction context.

Contextual Notes

Participants reference the molar volume and the stoichiometry of the reaction, but there are unresolved assumptions regarding the definitions of volume and the conditions under which the calculations are made.

markosheehan
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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.
 
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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
 

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