Question

Se ard in clor 190.09 grame de fier cu 12% impuritati. Ce cantitati de clorura de fier se formeaza ? Ce numar de moli de colr se consuma ?

Answer 1

100% subst pura - 12% imp. = 88%

p% = puritatea = mp x 100 /mip

mp = p% x mip /100 = 190.09 x 88/100

=> mp = 167,28 g Fe pur

167,28... n moli....... a g

2Fe + 3Cl2 => 2FeCl3

2x56...... 3............2x162,5

=> n = 167,28x3x73/2x56

=> n = 4,48 moli Cl2

=> a = 167,28x2x162,5/2x56

=> a = 485,41 g FeCl3