Question

Urgent!
Se ard 190,90g fier cu 12% impuritati cu clor.Se cere cantitatea de clorura de fier III obtinuta 2Fe+3 Cl=2FeCl3​

Answer 1

Răspuns:

MFe 56 g/mol;

MFeCl3 =1×56 + 3×35,5 = 162,5(g/mol).

100% - 12% = 88%.

100gFe.................88gFe pur

190,90gFe...........xg Fe pur

x = 190,90×88/100 = 167,992g Fe pur

2×56g ............. 2×162,5g

2Fe   +  3Cl2  =  2FeCl3

167,992g...................yg

y = 167,992×2×162,5/2*56= calculează