Question

A student has a 5.8g sample of a mixture of mercury (ll) oxide (which decomposes upon heating) and zinc (ll) oxide (which does not). She decomposes the mixture and isolates 154 mL water in the end (representing the displacement of oxygen). If the water temperature is 24 degrees C and the atmospheric pressure is 742 torr, what was the percent by mass of mercury oxide in the mixture?

Answer 1

2HgO ---> 2Hg + O2 ....(1)

Let n is moles of H2O.

Volume of H2O = (V) = 0.154 liters

Pressure (P) = 742 torr = 742/760 = 0.9763 atm

Temperature (T) = 24° C = 273 + 24 = 297 K

Gas constant (R) = 0.0821 L atm/mol K

Using ideal gas equation;

PV = nRT

n = PV/RT = 0.9763 * 0.154 / 0.0821 * 297 = 0.006166 moles

We know that,

2H2O ---> 2H2 + O2 .....(2)

2 moles H2O produces 1 mole O2

So, 0.006166 mole will produce = 0.006166/2 = 0.003083 moles of O2

Moles of O2 = 0.003083

Again from reaction (1);

1 mole O2 is produced by 2 mole HgO

So, 0.003083 mole O2 will be produced by = 2*0.003083 = 0.006166 moles of HgO

Mass of HgO = moles * molar mass = 0.006166 * 216.59 = 1.5533 grams

% of HgO = 1.5533 * 100 / 5.8 = 23.03 % ....Answer

Let me know if any doubts/answer is not matching.