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Physics 15 Online
OpenStudy (anonymous):

what volume of nitrous oxide can be produced

OpenStudy (anonymous):

what volume of nitrous oxide can be produced from the decomposition of 0.55 moles of ammonium nitrate

OpenStudy (anonymous):

\[NH _{4}NO _{3(aq)} \rightarrow N _{2}O _{(g)} + 2H _{2}O _{(l)}\] \[n(N _{2}O) = (1/3) * 0.55\]\[n(N _{2}O) = 0.18\] Using ideal gas law... 0.18 * 22.4 = ?L

OpenStudy (unklerhaukus):

i think you would you get \(0.55\) moles of \(N_2O\) because the stoichiometric ratio ammonium nitrate:Nitrous oxide is 1:1 \[V[L]=0.55[\text{mol}]\times22.4\left[{\frac{\text L}{\text{mol}} }\right] = \dots\]

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