Moles H2 required = 3 x 0.230=0.690
<span>moles H2 in excess = 0.750 -0.690 =0.060 </span>
<span>the ratio between N2 and NH3 is 1 : 2 </span>
<span>moles NH3 = 0.230 x 2 = 0.460</span>
the balanced equation for oxidation of isoborneol by naocl is written as follows
C10H18O + NaOCl ----> C10H16O + NaCl + H2O
isoborneol (C10H18O) react with NaOcl to form C10H16O NaCl and H2O