Question

Question #5cb6e

Answer 1

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`.