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.
I tried doing this using the ideal gas law, but whatever I did wasn't the right answer
To solve this problem, we can use the ideal gas law to calculate the number of moles of O2 gas produced during the decomposition of silver oxide. Then, we can calculate the percentage of silver oxide in the sample.
Given data:
Mass of the sample = 0.700 g
Volume of gas collected = 13.8 mL
Temperature = 27°C = 27 + 273 = 300 K
External pressure = 1.00 atm
Partial pressure of water vapor = 26.7 mmHg
Step 1: Convert the partial pressure of water vapor to atm.
Partial pressure of water vapor = 26.7 mmHg = 26.7/760 atm = 0.035 atm.
Step 2: Convert the volume of gas collected to liters.
Volume of gas collected = 13.8 mL = 13.8/1000 L = 0.0138 L.
Step 3: Calculate the total pressure of the gas.
Total pressure of the gas = external pressure - partial pressure of water vapor
Total pressure of the gas = 1.00 atm - 0.035 atm = 0.965 atm.
Step 4: Use the ideal gas law to calculate the number of moles of O2 gas.
PV = nRT
n = PV/RT
n = (0.965 atm)(0.0138 L) / (0.0821 L·atm/mol·K)(300 K)
n ≈ 0.000619 mol
Step 5: Determine the molar ratio between Ag2O and O2 from the balanced equation.
From the balanced equation: 2 Ag2O(s) → 4 Ag(s) + O2(g)
The molar ratio between Ag2O and O2 is 2:1.
Step 6: Calculate the number of moles of Ag2O.
Since the molar ratio between Ag2O and O2 is 2:1, the number of moles of Ag2O is half the number of moles of O2.
Number of moles of Ag2O = 0.000619 mol / 2 ≈ 0.0003095 mol
Step 7: Calculate the mass of Ag2O.
Since the molar mass of Ag2O is 231.74 g/mol:
Mass of Ag2O = number of moles of Ag2O × molar mass of Ag2O
Mass of Ag2O = 0.0003095 mol × 231.74 g/mol ≈ 0.0716 g
Step 8: Calculate the percentage of silver oxide in the sample.
Percentage of Ag2O = (mass of Ag2O / mass of the sample) × 100%
Percentage of Ag2O = (0.0716 g / 0.700 g) × 100%
Percentage of Ag2O ≈ 10.23%
Therefore, the percentage of silver oxide in the sample is approximately 10.23%.
To solve this problem, we need to calculate the number of moles of oxygen gas (O2) produced, and then relate it to the number of moles of silver oxide (Ag2O). Finally, we can express the percentage of silver oxide in the sample.
Let's break down the steps:
Step 1: Calculate the pressure of the oxygen gas (O2)
The total pressure is given as 1.00 atm, and the partial pressure of water vapor is given as 26.7 mm Hg. To make the units consistent, let's convert the partial pressure of water vapor from mm Hg to atm:
Pressure of water vapor (Pwater) = 26.7 mm Hg * (1 atm / 760 mm Hg) = 0.035 atm
The pressure exerted by the oxygen gas (Poxygen) is then:
Poxygen = Total pressure - Pressure of water vapor = 1.00 atm - 0.035 atm = 0.965 atm
Step 2: Convert the volume of gas collected to moles of oxygen gas (O2)
We're given that 13.8 mL of gas is collected. To convert this volume to moles, we need to use the ideal gas law equation:
PV = nRT
R represents the ideal gas constant (0.0821 L·atm/mol·K), T represents the temperature (in Kelvin), P is the pressure (in atm), and V is the volume (in liters).
However, since the collected gas is over water, we need to account for the vapor pressure of water at that temperature. In this case, we have to subtract the vapor pressure of water from the total pressure before applying the ideal gas law.
Step 3: Convert moles of oxygen gas to moles of silver oxide (Ag2O)
From the balanced chemical equation, we know that 1 mole of O2 corresponds to 2 moles of Ag2O.
Step 4: Calculate the mass of silver oxide (Ag2O)
We're given that the sample weighs 0.700 g. So we need to calculate the moles of silver oxide and then determine the mass using the molar mass of Ag2O (Molar mass of Ag2O = 231.74 g/mol).
Step 5: Calculate the percentage of silver oxide in the sample
Finally, we can express the percentage of silver oxide in the sample using the formula:
Percentage of silver oxide = (mass of Ag2O / mass of sample) * 100
By following these steps and using the given values, you should be able to solve the problem and find the correct answer.
2Ag2O ==> 4Ag + O2
PV = nRT
Ptotal = pO2 + pH2O
760 = pO2 + 26.7
pO2 = 760-26.7 and convert to atm.
Solve for n = mols O2 and don't forget to convert T to kelvin.
Then ? mols O2 x (2 mols Ag2O/1 mol O2) = ? mols O2 x 2/1 = ?? Ag2O
Convert mols Ag2O to grams. g = mols x molar mass
Then %Ag2O = (mass Ag2O/mass sample*100 = ?