Question

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.

Answer 1

Answer:

$\%Ag_2O=41.8\%$

Explanation:

Hello.

In this case, for the reaction:

$2 Ag_2O(s) \rightarrow 4 Ag(s) + O_2(g)$

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):

$PV=nRT\\\\n=\frac{PV}{RT}=\frac{(1.00atm-0.035atm)*0.0138L}{0.082\frac{atm*L}{mol*K}*300K} =5.41x10^{-4}molO_2$

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:

$m_{AgO_2}=5.41x10^{-4}molO_2*\frac{2molAgO_2}{1molO_2} *\frac{231.735gAgO_2}{1molAgO_2} \\\\m_{AgO_2}=0.251gAgO_2$

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

$\%Ag_2O=\frac{0.251g}{0.600g}*100 \%\\\\\%Ag_2O=41.8\%$

Best regards.

Answer 2

There is  42.5% of Ag2O in the sample.

The equation of the reaction is;

2 Ag2O(s) -------> 4 Ag(s) + O2(g)

Volume of the oxygen gas = 13.8 mL or 0.0138 L

Temperature of gas = 27°C + 273 = 300 K

Pressure of the gas = 1 atm - 0.035atm = 0.965 atm

Gas constant = 0.082 atm LK-1mol-1

From;

PV = nRT

n = PV/RT

n = 0.965 atm × 0.0138 L/0.082 atm LK-1mol-1 × 300 K

n = 0.000541 moles

If 2 moles of Ag2O yields 1 mole of oxygen

x moles of Ag2O yields 0.000541 moles of oxygen

x = 2 moles ×  0.000541 moles/ 1 mole

x = 0.0011 moles of Ag2O

Mass of Ag2O = 0.0011 moles × 232 g/mol = 0.255 g

Mass percentage of Ag2O =  0.255 g/0.600 g × 100/1

= 42.5%

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