-calculez numarul de moli de alchina, cu ecuatia de stare
pV= nRT
2X24,6= nx0,082x(27+273)
n= 2mol alchina
din ecuatie de aditie a apei, deduc alchina
1mol.............................1mol
CnH2n-2 + H20----.> CnH2nO
2mol.............................2mol------deci 2mol produs cantaresc 88g
2xM=88-----> M=44g/mol
dar
44g= 14n+16-------------------> n=5--------> CH≡ CH
CH≡CH+H2O---> CH3-CHO( ALDEHIDA ACETICA)
1 singur produs reactioneaza cu clorura diamino cuprica