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Sidana [21]
3 years ago
12

How many liters of water vapor can be produced if 8.9 liters of methane gas (CH4) are combusted, if all measurements are taken a

t the same temperature and pressure? Show all of the work used to solve this problem.
CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (g)
Chemistry
1 answer:
Delvig [45]3 years ago
8 0
We know, mole of gas in ideal conditions is 22.414 L
Here, CH4 volume = 8.9L

So, number of moles = 8.9 / 22.414 = 0.397 moles
Here, in balanced equation we have a ratio CH4 : H2O = 1:2.

So, mole so water = , 0.397*2= 0.794 moles
In liters it would be: 0.794 * 22.414 = 17.80 L

In short, Your Answer would be 17.80 Liters

Hope this helps!
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<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

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<h3>Percent yield in this case</h3>

In this case, you know:

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