$2H_2(g) + O_2(g) rarr 2H_2O(g)$
We started with $(5.0*g)/(32.00*g*mol^-1)=0.156*mol" dioxygen"$.
Given excess dihydrogen, this molar quantity thus gives $2xx0.156*mol$ water.
And thus, $2xx0.156*molxx18.01*g*mol^-1~=6*g$ water are evolved. Is this reaction exothermic or endothermic?