If 5.57 g of ag2o is sealed in a 75.0-ml tube filled with 760 torr of n2 gas at 28 ∘c, and the tube is heated to 310 ∘c, the ag2o decomposes to form oxygen and silver. part a what is the total pressure inside the tube assuming the volume of the tube remains constant?

Total pressure = 7293.2 torr or 9.60 atm

Explanation:

Total pressure = partial pressure of nitrogen + partial pressure of oxygen

The partial pressure due to nitrogen is determined using the equation of Gay-Lussac's law: P₁/T₁=P₂/T₂

P₁ = 760 torr = 1atm, T₁ = 28∘C = (273+28)K = 301k, P₂ = ?, T₂ = 310∘C =(310+273)K = 583K

P₂ = P₁ T₂/ T₁

P₂ = 760 * 583 / 301 = 1472.03 torr

The pressure due to Oxygen gas produced is calculated thus:

Balanced equation of the decomposition of Ag₂O at s.t.p. is as follows;

2Ag₂O > 4Ag + O₂(g)

2 moles of Ag₂O produces 1 mole of O₂

molar mass of Ag₂O = (2*108 + 16)g = 232g/mol

molar volume of gas at s.t.p. = 22.4L

2*232g i.e. 464g of Ag₂O produces 22.4L of O₂

5.57g of Ag₂O will produce 5.57g*22.4L/464g = 0.269L or 269mL of O₂

Using the General gas equation  P₁V₁/T₁=P₂V₂/T₂

P₁ = 1atm = 760 torr, V₁ = 269mL, T₁=273K, P₂ = ?, V₂= 75mL, T₂ = 583K

P₂ = P₁V₁T₂/V₂T₁

P₂ = 760*269*583 / 75*273

P₂ = 5821.17 torr

Total pressure = (1472.03 + 5821.17) torr

Total pressure = 7293.2 torr or 9.60 atm

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