Percent Yield?

Question

Mercury used to be prepared in the laboratory by heating mercuric oxide

2HgO(s) 2Hg(l) + O2(g)

If 1.87 g of mercuric oxide were heated, what would be the theoretical yield of mercury?

Theoretical yield = ? g Hg

Suppose 1.09 g of mercury was actually collected: what is the percent yield?

Percent yield = %

Answer 1

2HgO(s) ->2Hg(l) + O2(g)

Hg - 200.59 g/mol
O2 - 31.98 g/mol
HgO- 216.58 g/mol

Since 2 mol of HgO yield 2 moles of Hg

1.87/(2x216.58 )=x/(2x 200.59 )
x= (2x 200.59 )x1.87/(2x216.58 )
x= ( 200.59 )x1.87/(216.58 )
x=1.73g (theoretical yeld(ty))

% ty=(1.73/1.87)x100= 92.5%

Practica yeld

%py=(1.09/1.87)x100=58.3%