From the follortring enthalpy changes,

$2PbS(s)+3O_2(g)2$PbO$(s)+2S{O}_2(g),H=-827\mathrm{.}0kJ$

PbO$(s)+C(s)Pb(s)+CO(g),H=+106\mathrm{.}8kJ$

calculate the value of $H$ when $1\mathrm{.}55$mol of $PbS$ reacts to form lead in the following reaction:

$2PbS(s)+3O_2(g)+2C(s)2Pb(s)+2CO(g)+2S{O}_2(g).$ Is the reaction endothermic or exothermic?