Question

For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.
2HgO(s)→2Hg(l)+O2(g)

A. 2.55g HgO

B. 6,89g HgO

C. 1.50kg H_gO

D. 3.95mg H_gO   

Answer 1

Solution :

The given reaction is :

2HgO(s) = 2Hg (l) + O2(g)

Thus, 2 mol HgO produces 1 mol of O2

A) 2.55 g HgO :

Number of mol of HgO = Mass /molar mass

= 2.55 g / 216.59 g mol-1 = 0.0118 mol

Hence, mol of O2 = 0.0118 mol /2 = 0.0059 mol

Mass of O2 = number of mol x molar mass

= 0.0059 mol x 32 g mol-1 = 0.189 g

B) 6.89 g HgO:

Number of mol = 6.89 g /216.59 gmol-1 = 0.0318 mol

Thus, mol of O2 = 0.0318 / 2 = 0.0159 mol

Mass of O2 = 0.0159 mol x 32 g mol-1 = 0.509 g

C) 1.50 Kg HgO

Number of mol = 1500 g / 216.59 gmol-1 = 6.926 mol

Mol of O2 = 6.926 /2 = 3.463 mol

Mass of O2 = 3.463 mol x 32 g mol-1 = 110.82 g

D) 3.95 mg HgO = 0.00395 g

Moles of HgO = 0.00395 g / 216.59 g mol-1 = 1.82 x 10^-5 mol

Mol of O2 = 1.82 x 10^-5 /2 = 9.1 x 10^-6 mol

Mass of O2 = 9.1 x 10^-6 mol x 32 g mol-1 = 2.91 x 10^-4 g

= 0.0291 mg