A sample of nitrogen gas is produced in a reaction and collected under water in a graduated cyliner. The temperature is 26.3 oC
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I found the attached image with the same statement of your question and think it may be very useful for you that I use it to show how to answer this question (furthermore I think it may be the same reaction that you forgot to include).
As you can see, there are one image on the left side and other image on hte right side of the figure.
Those images contains drawings that represent molecules and a legend that permit you to distinguish the kind of atoms in each molecule.
Using that, you can indicate the chemical reaction as the transformation of the molecules on the left side onto the molecules on the right side:
Left side:
3 molecules of CH4 and 3 molecules of N2Cl4
Right side:
3 molecules of CCl4, 3 molecules of N2 and 6 molecules of H2
That is represented as:
3CH4 + 3 N2Cl4 -----> 3 CCl4 + 3N2 + 6H2
And that is the balanced chemical equation for the reaction shown in the figured attached.
I hope this is useful for you..
As you can see, there are one image on the left side and other image on hte right side of the figure.
Those images contains drawings that represent molecules and a legend that permit you to distinguish the kind of atoms in each molecule.
Using that, you can indicate the chemical reaction as the transformation of the molecules on the left side onto the molecules on the right side:
Left side:
3 molecules of CH4 and 3 molecules of N2Cl4
Right side:
3 molecules of CCl4, 3 molecules of N2 and 6 molecules of H2
That is represented as:
3CH4 + 3 N2Cl4 -----> 3 CCl4 + 3N2 + 6H2
And that is the balanced chemical equation for the reaction shown in the figured attached.
I hope this is useful for you..
It takes 21.3 days
<h3>Further explanation</h3>Given
5 hr = 8 kg Alcohol
Required
Days to consume 1000 kg of glucose
Solution
Alcoholic fermentation⇒ glucose produces ethanol and carbon dioxide,
C₆H₁₂O₆ → 2 C₂H₅OH + 2CO₂
mol ethanol :
moles of glucose to produce 108.7 moles ethanol :
54.35 moles = 5 hours
moles of 1000 kg of glucose :
So for 5555.5 moles, it takes :