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)

6.27 g HgO

1.14 kg HgO

Answer 1

2HgO(s) ------> 2Hg(l) + O2(g)

2 moles of HgO gives 2 moles of Hg and 1 mole of O2.

number of moles mole = (wt. in g / molecular wt. in g/mol.)

6.27 g HgO = (6.27 g) / (216.591 g/mol) = 0.02895 moles

Thus we know that 2 moles of HgO gives 1 mole O2.

so 0.02895 moles of Hg will give 0.01447 moles of O2.

Molecular wt. of O2 = 32 g/mol

Hence, 0.01447 moles of O2 = (0.01447 moles) * (32 g/mole) = 0.46304 g of O2. Thus 6.27 g of HgO will produce 0.46304 g of O2.

Now similarly, 1.14 kg of HgO = 1140 g of HgO.

1140 g HgO = (1140 g) / (216.591 g/mol) = 5.2634 moles

so 5.2634 moles of Hg will give 2.6317 moles of O2.

2.6317 moles of O2 = (2.6317 moles) * (32 g/mole) = 84.2144 g of O2. Thus 1.14 Kg (or 1140 g) of HgO will produce 84.2144 g of O2.