If 49.8g of MgO(s) is reacted with 14.6g of H2S, (a) what is the limiting reagent, and (b) what is the mass of the leftover reagent?
The balanced equation is:
MgO(s) + H2S(g) --> MgS(s) + H2O(l)
The molar mass of MgO is 40.30 g/mol
The molar mass of H 2S is 34.08 g/mol
Show your work and round to the correct number of significant figures.