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Setler79 [48]
3 years ago
11

A 6.0 mol sample of C3H8(g) and a 20. mol sample of Cl2(g) are placed in a previously evacuated vessel, where they react accordi

ng to the equation above. After one of the reactants has been totally consumed, how many moles of HCl(g) have been produced?
Chemistry
2 answers:
Vika [28.1K]3 years ago
6 0

Answer:

20

Explanation:

20 mol of cl2 / 4  cl2 = 5 for the equivalence

5* 4 cl = 20

because one in completeely used, cl2 is the limiting reagent,

gladu [14]3 years ago
3 0
2 CH₃CH₂CH₃ + 2 Cl₂ →  CH₃CH(Cl)CH₃ + CH₃CH₂CH₂Cl + 2 HCl
so according to balanced equation we notice that:
2 mole of propane reacts with 2 moles of Cl₂, so 6 moles of propane will need to react completely with 6 moles Cl₂ (This mean Cl₂ present in excess and propane is the Limiting reagent)
2 mole C₃H₈ gives 2 moles HCl
6 moles C₃H₈ gives ? moles HCl
by cross multiplication we have 6 moles of HCl formed  
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Answer:

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Where x represent the value of interest on this case. And solving for the value of x we have:

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And in order to find the concentration we can use a figure called the "O2

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So then the final answer for this case would be an increase of 7%

Explanation:

For this case we know that a man with normal lungs have an arterial Po2 os 40 mm Hg.

Then we know that this man take an overdose of barbiturates thats halves his aveolar ventilation without changing his metabolism

We also know that the respiratory exchange ratio is 0.8 or 8/10

0.8 = \frac{8}{10}= \frac{4}{5}

And we want to find hor much does his inspired oxyden concentration % have to increased to return his avelolar Po2 to the original level.

On this case we can apply a proportion rule and we have this:

\frac{4}{5} = \frac{40}{x}

Where x represent the value of interest on this case. And solving for the value of x we have:

x = 40 * \frac{5}{4}= 50 mm Hg

So then the new arterial pressure needs to be now 50 mm Hg to mantain the original level.

And in order to find the concentration we can use a figure called the "O2

-CO2  diagram showing a ventilation-perfusion ratio line. " and when we use this graph to calculate the pressure of Co2 for PO2= 40 mmHg and for Po2=50 mm Hg we got and increase of 0.07 or 7%  

So then the final answer for this case would be an increase of 7%

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7 0
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