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.

Explanation:

Hello.

In this case, for the reaction:

As 13.8 mL (0.0138 L) are collected at 27 °C and 1.00 atm, we can compute the yielded moles of oxygen via the ideal gas equation as shown below, considering that the pressure of oxygen is equal to the total pressure minus the vapor pressure of water (26.7 mmHg=0.035atm):

Next, via the 1:2 mole ratio between oxygen and silver oxide (molar mass = 231.735  g/mol) in the chemical reaction, the mass of pure silver oxide turns out:

In such a way, the percent by mass of silver oxide in the 0.600-g sample is:

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answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :