Question

Masa impură = 190,9 g Fe
P= 88%
a) m = ? FeCl3
b) n = ?

Answer 1

p% = puritatea = mpx100/mip

 => mp = mipxp%/100 = 190,9 x 88/100= 168 g Fe pur

168 g ...................... m g

2Fe  +  3Cl2  => 2FeCl3

2x56........................ 2x(56+35,5x3)

=> m = 168x2x162,5/2x56 = 487,5 g FeCl3

niu = masa subst. / M.moleculara subst.

=> niu = 487,5 / 162,5 = 3 moli FeCl3