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) 6.44 HgO 1.05 Kg HgO 3.83 mg HgO   

6.44 HgO

1.05 Kg HgO

3.83 Mg HgO

Answer 1

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

2 moles of HgO react to give 1 mole of O2

1) 6.44 g HgO

Number of moles = amount in g / molar mass

no. of mole of HgO = 6.44 g /216.6 g / mole

= 0.0297 mole HgO

Now calculate the moles of O 2 as follows:

0.0297 mole HgO * 1 moles O2/ 2 mole HgO

= 0.01485 moles O2

Amount of O2 = number of moles * molar mass

= 0.01485 moles O2 * 31.98 g/ moles

= 0.475 g O2

2)1.05 kg or 1050 g HgO

Number of moles = amount in g / molar mass

no. of mole of HgO = 1050 g /216.6 g / mole

= 4.85 mole HgO

Now calculate the moles of O 2 as follows:

4.85 mole HgO * 1 moles O2/ 2 mole HgO

= 2.425 moles O2

Amount of O2 = number of moles * molar mass

= 2.425 moles O2 * 31.98 g/ moles

= 77.55 g O2

3) 3.83 mg or 0.00383 g HgO

Number of moles = amount in g / molar mass

no. of mole of HgO = 0.00383   g /216.6 g / mole

= 1.77*10^-5 mole HgO

Now calculate the moles of O 2 as follows:

1.77*10^-5 mole HgO * 1 moles O2/ 2 mole HgO

= 8.84*10^-6 moles O2

Amount of O2 = number of moles * molar mass

= 8.84*10^-6 moles O2 * 31.98 g/ moles

= 2.83*10^-4 g O2

= 0.283 mg O2