Answer:
[tex]n_{N_2}=0.00369molN_2[/tex]
Explanation:
Hello there!
In this case, by firstly setting up the described chemical reaction, we are able to write:
[tex]2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(g)[/tex]
Thus, given the pressure (convert to atm), volume and temperature (convert to K) we can calculate the moles of NO:
[tex]n=\frac{PV}{RT}\\\\n=\frac{200.00/101.325atm*0.090L}{0.08206\frac{atm*L}{mol*K}*(20.0+273.15)K} \\\\n=0.00738molNO[/tex]
Next, we use the 2:1 mole ratio of NO to N2 to obtain the moles of nitrogen gas that will be produced:
[tex]n_{N_2}=0.00738molNO *\frac{1molN_2}{2molNO} \\\\n_{N_2}=0.00369molN_2[/tex]
Regards!
