Răspuns :
240 4*22.4
HC≡CH+2[Ag(NH3)2]OH-> AgC≡CAg + 4NH3+ 2H2O
0.5 x
MC2Ag2=12*2+108*2=24+216=240g/mol
Se calculeaza volumul de NH3 de pe reactie
x=0.5*4*22.4/240=0.1866 l NH3
V c.n. = 22,4 l/mol
M Ag2C2 = 240 g/mol
240 g Ag2C2.......4*22,4 l NH3
0,5 g Ag2C2...............x
x = 0,18666 l NH3
M Ag2C2 = 240 g/mol
240 g Ag2C2.......4*22,4 l NH3
0,5 g Ag2C2...............x
x = 0,18666 l NH3