mass Ag = 59.5 g
Given
64 g of Ag2O
Required
Mass of Ag
Solution
Proust stated the Comparative Law that compounds are formed from elements with the same Mass Comparison so that the compound has a fixed composition of elements
mass of Ag in Ag2O :
= ((2 x Ar Ag)/molar mass Ag2O )x mass Ag2O
[tex]\tt mass~Ag=\dfrac{2.Ar~Ag}{MW~Ag_2O}\times mass~Ag_2O[/tex]
Ar Ag : 107,8682 g/mol
MW Ag₂O = 231,735 g/mol
Input the value :
mass Ag = ( (2 x 107,8682 g/mol )/ 231.735) x 64 g
mass Ag = 59.5 g
Or we can use reaction
2Ag₂O ⇒ 4Ag + O₂
mol Ag₂O :
= 64 : 231.735
= 0.276
mol Ag :
= 4/2 x moles Ag₂O
= 4/2 x 0.276
= 0.552
mass Ag :
= 0.552 x 107,8682 g/mol
= 59.54 g
