Răspuns :
HCOOH + CH3OH------> HCOOCH3+ H2O
M,ester= 60g/mol
M,acid=46g/mol
M,alcool= 32g/mol-
-calculez moli ester obtinuti
n=m/M= 9,6g/60g/mol= 0,16mol
-calculez moli de acid care intra in reactie, tinand cont de randament
n,acid= 100x0,16/80=0,2mol
-calculez moli alcool din raportul molar dat
n,acid/n,alcool= 1/3-------> n,alcool= 0,6mol
-calculez masa de amestec introdus in reactie
m-=0,2x46 g+ 0,6x32=...............................calculwaza !!!!!
M,ester= 60g/mol
M,acid=46g/mol
M,alcool= 32g/mol-
-calculez moli ester obtinuti
n=m/M= 9,6g/60g/mol= 0,16mol
-calculez moli de acid care intra in reactie, tinand cont de randament
n,acid= 100x0,16/80=0,2mol
-calculez moli alcool din raportul molar dat
n,acid/n,alcool= 1/3-------> n,alcool= 0,6mol
-calculez masa de amestec introdus in reactie
m-=0,2x46 g+ 0,6x32=...............................calculwaza !!!!!