Considering this balanced chemical equation, how many grams of HgO will be produced when 44 g of Hg react with excess O2? 2Hg (l) + O2 (g) --> 2HgO (s)

A 28 g
B 44 g
C 48 g
D 96 g

Also, how would you go to solve this problem? Thanks!

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)

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year