Question

1) a)Ce cantitate de sulfat de cupru reactioneaza cu 11,2g KOH?
b) aflati nr de moli de KOH
2) cate grame de acid fosforic reactioneaza cu 0,8 moli hidroxid de Ca.
Multumesc

Answer 1

1. a) 2KOH + CuSO4 = K2SO4 + Cu(OH)2

MKOH=AK+AO+AH=39+16+1=56g/mol

MCuSO4=ACu+AS+4AO=64+32+4*16=160g/mol

2*56g KOH....................160g CuSO4
11,2g KOH......................x=11,2*160/2*56= Calculeaza !!!

b) n=m/M=11,2/56=0,2 moli KOH

2. 2H3PO4 + 3Ca(OH)2 = Ca3(PO4)2 + 6H2O

MH3PO4=98g/mol

MCa(OH)2=74g/mol

2 moli H3PO4............3*74g Ca(OH)2
0,8 moli H3PO4............x=88,8g