A 0.700 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

1. Convert the volume of gas collected to moles:
PV = nRT
n = PV/RT = (1.00 atm - 0.035 atm) * (13.8 mL/1000 mL/L) / (0.0821 L*atm/K*mol * (27°C + 273.15)K) = 0.00058 mol

(The pressure of water vapor over the water is 26.7 mmHg, which is equivalent to 0.035 atm. This value is subtracted from the atmospheric pressure to get the partial pressure of oxygen.)

2. Use stoichiometry to find the number of moles of Ag2O that decomposed:
2 Ag2O -> 4 Ag + O2
1 mol Ag2O gives 1/2 mol O2
0.00058 mol O2 came from the decomposition of 0.00058/1/2 = 0.00116 mol Ag2O

3. Calculate the percentage of Ag2O in the sample:
0.00116 mol Ag2O is present in 0.700 g of the sample.
1 mol Ag2O has a mass of (108 + 32 + 16*4) g/mol = 231 g/mol
0.00116 mol Ag2O has a mass of 0.268 g
The percentage of Ag2O in the sample is (0.268 g/0.700 g) * 100% = 38.4%

Therefore, the answer is choice C (35.9%).

A 0700 G Sample Containing Ag2O and Inert Material Is Heated,causing

A 0700 G Sample Containing Ag₂O and Inert Material Is Heated,causing