chemistry law gas

Question

How many grams of Ag 2 O Ag 2 ​ O Subscript 2 Baseline upper O are required to produce 1.15 L oxygen gas via decomposition at 102.3 kPa and 175°C? Use the ideal gas law to calculate moles of O₂. Use the mole ratio to calculate moles Ag₂O. Use molar mass to calculate grams. Don't forget to convert to atm and K!

Answer 1

Balanced equation for decomposition of Ag2O is...

2Ag₂O ==> 4Ag + O₂(g)

It says to use the ideal gas law:

PV = nRT

n = PV/RT

Solving for n (moles of O₂) we have...

n = moles = ?

P = pressure = 102.3 kPa converted to atm is 102.3 kPa x 1 atm / 101.325 kPa = 1.009 atm

V = volume = 1.15 L

R = 0.0821 Latm/Kmol

T = temp in K = 175 + 273 = 448K

n = (1.009)(1.15)/(0.0821)(448)

n = 0.0315 moles O₂

Mole ratio O₂ to Ag₂O = 1 mol O₂ to 2 mol Ag₂O (see balanced equation)

moles Ag₂O required = 0.0315 mol O₂ x 2 mol Ag₂O / mol O₂ = 0.0630 mols Ag₂O needed

Molar mass Ag₂O = 231.7 g / mole

grams Ag₂O needed = 0.0630 mols x 231.7 g / mol = 14.6 g Ag₂O needed

Answer 2

Ag₂O ⇒ 2Ag + (1/2)O₂

1.15 liters, 102.3 kPa (175+273)K Use ideal gas law to solve for moles of O₂ n = PV/(RT)

You can either use 8.314 for R or convert to atm (kPa/101.3) and use .0821 for R

Moles O₂(1 mole Ag₂O/1/2 mole of O₂)( 231.7 g/ mole Ag₂O) = g Ag₂O (stop before last conversion if you want moles of silver oxide.

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