Complete Question:
1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid mercury(II) oxide (HgO) into liquid mercury and oxygen gas.
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2. Suppose 59.0 mL of dioxygen gas are produced by this reaction, at a temperature of 90.0°C and a pressure of exactly 1 atm. Calculate the mass of mercury (II) oxide that must have reacted.
Answer:
1. 2HgO(s) → 2Hg(l) + O₂(g)
2. 0.8664 g
Explanation:
1. The reaction will be:
HgO(s) → Hg(l) + O₂(g)
To balance the equation, all the elements must have the same amount at both sides of the equation, so HgO and Hg must be multiplied by 2:
2HgO(s) → 2Hg(l) + O₂(g)
2. By the ideal gas law, we can find how many moles of O₂ was produced:
PV = nRT
Where P is the pressure (1 atm), V is the volume (59.0 mL = 0.059 L), n is the number of moles, R is the gas constant (0.082 atm.L/mol.K), and T is the temperature (90.0°C = 363 K).
1*0.059 = n*0.082*363
29.766n = 0.059
n = 0.002 mol
By the stoichiometry of the reaction:
2 moles of HgO ------------------ 1 mol of O₂
x ------------------ 0.002 mol
By a simple direct three rule:
x = 0.004 mol of HgO
The molecular mass of HgO is 216.59 g/mol. The mass is the number of moles multiplied by the molecular mass:
m = 0.004 * 216.59
m = 0.8664 g
