Answer:
[tex]6.68~g~NO_2[/tex]
Explanation:
We have to start with the combustion reaction:
[tex]NO~+~O_2~->~NO_2[/tex]
Then we can balance the reaction:
[tex]2NO~+~O_2~->~2NO_2[/tex]
If we want to find the theoretical yield, we have to calculate the amount of [tex]NO_2[/tex]. To do this, we have to first convert the 4.36 g of [tex]NO[/tex] to moles [tex]NO[/tex] (using the molar mass 30 g/mol), then we have to convert from moles of [tex]NO[/tex] to moles of [tex]NO_2[/tex] (using the molar ratio) finally, we have to convert from moles of [tex]NO_2[/tex] to grams of [tex]NO_2[/tex] (using the molas mass 46 g/mol), so:
[tex]4.36~g~NO\frac{1~mol~NO}{4.36~g~NO}\frac{2~mol~NO_2}{2~mol~NO}\frac{46~g~NO_2}{1~mol~NO_2}=6.68~g~NO_2[/tex]
I hope it helps!