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A #21.3*g# mass of #"nitrous oxide"#, #N_2O#, is oxidized by dioxygen gas to #NO_2(g)#. How much dioxygen is required?

1 Answer

Dec 14, 2016

#N_2O(g) + 3/2O_2(g)rarr2NO_2(g)#

Explanation:

#"Moles of"# #N_2O# #=# #(21.3*g)/(44.02*g*mol^-1)=0.483*mol#

And for stoichiometric reaction we need,

#0.483*molxx3/2xx32.00*g*mol^-1=??*g# #"dioxygen gas"#.

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