Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.
2 S + 3 O₂ → 2 SO₃
The stoichiometric calculations is as follows:
7 g S * 1 mol/32.06 g S = 0.218 mol S Moles O₂ needed = 0.218 mol S * 3 mol O₂/2 mol S = 0.3275 mol O₂ Since the molar mas of O₂ is 32 g/mol, Mass of O₂ needed = 0.3275 mol O₂ * 32 g/mol = 10.48 g O₂
The question is incomplete. The complete question is :
In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an explanation that tells how much NaOH needs to be added to a beaker of HCl to cause the color to change. Your explanation can be something like: The color change will occur when [some amount] of NaOH is added because the color change occurs when [some condition]. The goal for your explanation is that it describes the outcome of this example, but can also be used to predict the outcome of other examples of this phenomenon. Here's an example explanation: The color of the solution will change when 40 ml of NaOH is added to a beaker of HCl because the color always changes when 40ml of base is added. Although this explanation works for this example, it probably won't work in examples where the flask contains a different amount of HCl, such as 30ml. Try to make an explanation that accurately predicts the outcome of other versions of this phenomenon.
Solution :
Consider the equation of the reaction between NaOH and
NaOH (aq) + HCl (aq) → NaCl(aq) +
The above equation tells us that of reacts with of .
So at the equivalence point, the moles of NaOH added = moles of present.
If the volume of the taken = mL and the conc. of = mole/L
The volume of NaOH added up to the color change = mole/L
Moles of taken = moles.
The color change will occur when the moles of NaOH added is equal to the moles of taken.
Thus when
or when
or mL of NaOH added, we observe the color change.
Where are the volume and molarity of the taken.
is the molarity of NaOH added.
When both the NaOH and are of the same concentrations, i.e. if , then
Or the 40 mL of will need 40 mL of NaOH for a color change and
30 mL of would need 30 mL of NaOH for the color change (provided the concentration )
A gas with a vapor density greater than that of air, would be most effectively displaced out off a vessel by ventilation.
The two following principles determine the type of ventilation: Considering the impact of the contaminant's vapour density and either positive or negative pressure is applied.
Consider a vertical tank that is filled with methane gas. Methane would leak out if we opened the top hatch since its vapour density is far lower than that of air. A second opening could be built at the bottom to greatly increase the process' efficiency.
A faster atmospheric turnover would follow from air being pulled in via the bottom while the methane was vented out the top. The rate of natural ventilation will increase with the difference in vapour density. Numerous gases that require ventilation are either present in fairly low concentrations or have vapor densities close to one.