Question

How many moles of NO(g) can be produced in the reaction of 3.00 mol NH3(g) and 4.00 mol O2(g) ? 

4 NH3(g) + 5 O2(g) = 4 NO(g) + 6 H2O(l)

Answer 1

3 moles of NH$_3$ requires $\frac{3}{4}×5=3.75\>moles$ of $O_2$

4 moles of O₂ requires $\frac{4}{5}×4=3.2$ moles of NH₃

Limiting reagent is NH₃

Moles of NO $=$ moles of NH₃

$= 3$ moles