Question

care este mas hidroxidului de Al(OH)3 format la interactiunea sulfatului de aluminiu cu 7,4 gde var stins Ca(OH)2

Answer 1

7,4g. m-?
Al2(SO4)3+3Ca(OH)2➡2Al(OH)3+3CaSO4
_______ _______
3•74=222g. 2•78=156g
M |Ca(OH)3|=40+2•16+2•1=74g/mol
M |Al(OH)3|=27+3•16+3•1=78g/mol

7,4g... m
222g...156g
m |Al(OH)3|=7,4g•156g/222g=5,2g