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How much Ag can be produced if one is given 64 g of Ag2O?

Sagot :

mass Ag = 59.5 g

Further explanation

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