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Chemistry 11 Online
OpenStudy (goalieboy):

Will post in attachments

OpenStudy (goalieboy):

here is the next one that I need help with @Michele_Laino

OpenStudy (michele_laino):

first step: we have to compute the number of moles of water, so what is: \[n = \frac{{90}}{{18}} = ...?\] since \(18\) is the molecular weight of water

OpenStudy (goalieboy):

that would be 5

OpenStudy (michele_laino):

correct! Now, looking at the right side of the chemical reaction, we note that number of moles of oxygen are hal the number of moles of water, so the moles of produced oxygen are: \[{n_1} = \frac{5}{2} = ...?\]

OpenStudy (goalieboy):

that is 2.5

OpenStudy (michele_laino):

oops.. are half... correct!

OpenStudy (michele_laino):

finally, one mole of oxygen occupies \(22.4\) liters, so \(2.5\) moles of oxygen will occupy: \(V=22.4 \cdot 2.5=...?\)

OpenStudy (goalieboy):

56

OpenStudy (michele_laino):

that's right!

OpenStudy (goalieboy):

so do I need to close this question and open up a new question?

OpenStudy (michele_laino):

yes!

OpenStudy (goalieboy):

ok

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