If 0.740 g of $O_3$ reacts with 0.670 g of $NO$, how many gram of $N{O}_2$ will be produced? 

$O_3+NO\to N{O}_2+O_2$

$0.74$ g of $O_3$ contains $0.74/48$ moles of $O_3$

That is, $0.0154$ mol of $O_3$

$0.67$ g of $NO$ contains $0.67/30$ moles of $NO$

That is, $0.0223$ mol of $NO$

$NO$  is in excess and $O_3$ is the limiting reagent = $0.0223 – 0.0154$

= $0.007$moles

Grams of $N{O}_2$ formed = molecular weight $\times$ moles

$=46g/mol\times 0.0154mol$

 $=0.71g$

Therefore, Option (A) is correct.