A 0.600 g sample containing Ag₂O and inert material is heated,causing the silver oxide to decompose according to the following equation:
2 Ag₂O(s) → 4 Ag(s) + O₂(g)
If 13.8 mL of gas are collected over water at 27°C and 1.00 atm external pressure,what is the percentage of silver oxide in the sample? The partial pressure of water is 26.7 mm Hg at 27°C.

Question

Quizplus · Accepted Answer

First, we need to calculate the moles of gas collected using the ideal gas law:

PV = nRT

n = (PV)/(RT)
n = ((1.00 atm)(0.0138 L))/(0.08206 L·atm/mol·K)(300 K)
n = 0.000668 mol

Next, we can use the balanced equation to find the moles of Ag2Sb12O:

2 Ag2Sb12O --> 4 Ag + O2
1 mol Ag2Sb12O produces 1/2 mol O2
0.000668 mol O2 * (1 mol Ag2Sb12O / 1/2 mol O2) = 0.001336 mol Ag2Sb12O

Finally, we can calculate the percentage of Ag2Sb12O in the sample:

mass of sample = mass of Ag2Sb12O + mass of inert material
mass of Ag2Sb12O = n × MM × (molar ratio of Ag2Sb12O in sample) 
MM of Ag2Sb12O = (2 × 107.87 g/mol) + 121.76 g/mol + 15.99 g/mol × 12 = 2383.3 g/mol
molar ratio of Ag2Sb12O in sample = 1 mol Ag2Sb12O / 4 mol Ag
mass of Ag2Sb12O = 0.001336 mol × 2383.3 g/mol × (1/4) = 0.0799 g

% Ag2Sb12O = (mass of Ag2Sb12O / total mass of sample) × 100%
% Ag2Sb12O = (0.0799 g / 0.600 g) × 100%
% Ag2Sb12O = 13.32%

Since the question asks for the percentage of silver oxide (Ag2O) in the sample, we need to note that Ag2O was converted to Ag during the decomposition reaction. Therefore, the percentage of Ag2O in the sample is twice the percentage of Ag2Sb12O:

% Ag2O = 2 × % Ag2Sb12O
% Ag2O = 2 × 13.32%
% Ag2O ≈ 26.6%

The closest answer choice is C, 41.8%, but none of the answer choices are exact.

A 0600 G Sample Containing Ag2O and Inert Material Is Heated,causing

A 0600 G Sample Containing Ag₂O and Inert Material Is Heated,causing