A) Balance the equation below.
Hg(ONC)2 > Hg + N2 + CO

b.) After decomposition, if you are left with 50.2g of CO, how many grams of mercury fulminate did you start with? (Hint: This problem asks about how many grams of reactant we start with, you can do the problem the same way as if CO was the reactant and Hg(ONC)2 was the product.)

254.7 g

Explanation:

Hg(ONC)2 > Hg + N2 + 2CO

In every problem of stoichiometry, we must commence with writing the balanced reaction equation as a guide to our calculation. Once we have written the reaction equation, we can now proceed with the solution.

Amount of CO produced= mass of CO produced/ molar mass of CO

Molar mass of CO = 28 gmol-1

Mass of CO produced= 50.2g

Amount of CO produced= 50.2g/28gmol-1

Amount of CO produced= 1.79 moles of CO

From the reaction equation:

1 mole of Hg(ONC)2 produced 2 moles of CO

x moles of Hg(ONC)2 will produce 1.79 moles of CO

x= 1.79 × 1 / 2

x= 0.895 moles of Hg(ONC)2

Molar mass of Hg(ONC)2 = 284.624 g/mol

Mass of Hg(ONC)2 = number of moles × molar mass = 0.895 moles × 284.624 g/mol

Mass of Hg(ONC)2 = 254.7 g

catya that's not even a b or c