Answer:
48g
Explanation:
2Hg(l) + O₂(g) => 2HgO
44g excess ?
44g/200.59g·mol⁻¹ = 0.219 mole Hg(l) = 0.219 mole HgO(s) produced.*
*Since coefficients are equal for Hg and HgO then the moles of HgO produced is the same as moles Hg used.
Convert 0.219 mole HgO to grams => 0.219mol x 216.95g/mol = 47.512g HgO ≅ 48g (2 sig figs)