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 HgO
D. 3.95mg HgO
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
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