Question #5cb6e

1 Answer
Jun 8, 2017

Answer:

We need (i) a stoichiometric equation:

#NO(g) + 1/2O_2(g) rarr NO_2(g)# and gets approx. #1*g# #NO_2#.

Explanation:

And (ii) we work out equivalent quantities of #NO(g)# and #"dioxygen gas"#......

#"Moles of NO"=(0.66*g)/(30.01*g*mol^-1)=0.0212*mol.#

#"Moles of dioxygen"=(0.58*g)/(32.00*g*mol^-1)=0.0181*mol.#

Given the stoichiometric equation, dioxygen is in EXCESS, and given quantitative reaction (which is not at all likely), we thus make #0.0212*mol# of #NO_2(g)#, a mass of #0.0212*molxx46.01*g*mol^-1=0.98*g#.