This is a limiting reagent (LR) problem. I know that because amounts are given for BOTH reactants.
1. Convert grams Hg to mols. mols = grams/molar mass.
2. Do the same for g oxygen to mols.
3a. Using the coefficients in the balanced equation, convert mols Hg to mols HgO.
3b. Do the same to convert mols O2 to mols HgO
3c. It is likely that the two values from 3a and 3b will not agree which means one of them is wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
4. Now use the smaller value mols HgO and convert to grams. g = mols x molar mass
NOTE: The correct way to show the name is mercury(II) oxide.(not ii)
2Hg+O2=2HgO
if 50.0 grams of mercury reacts with 50.0 grams of oxygen gas, how many grams of mercury(ii)oxide are produced?
1 answer
1 answer