C6H6+Br2= C6H5Br+HBr m= plotnost *V( obyem )=0,88* 8,86=7,80 gramm C6H6 n ( kolichestva veshestva )= m/M UZNAEM , CHTO U NAS V IZBITKE , A CHTO V NEDOSTATKE . n( C6N6)= 7,80/78 =0.1 mol( NEDOSTATOK ) n (Br2)= 24/160=0.15 ( IZBITOK ). NAXODIM x gramm po benzolu C6H6 : 7,80 GR SOOTVETSTVUET-x grammC6H5Br
C6H6+Br2= C6H5Br+HBr m= plotnost *V( obyem )=0,88* 8,86=7,80 gramm C6H6 n ( kolichestva veshestva )= m/M UZNAEM , CHTO U NAS V IZBITKE , A CHTO V NEDOSTATKE . n( C6N6)= 7,80/78 =0.1 mol( NEDOSTATOK ) n (Br2)= 24/160=0.15 ( IZBITOK ). NAXODIM x gramm po benzolu C6H6 : 7,80 GR SOOTVETSTVUET-x grammC6H5Br
78 MR C6H6 sootvetstvuet 157 Mr (C6H5Br) OTSYUDA X= 7,80*157 /78= 15,7 gramm (C6H5Br) teper naydem vixod .to chto nashi 15,7 gramm C6H5Br - ETO 100%
14,13 gramm <to chto dano po uslovii zadachi - x% VIXODA OTSYUDA x%= 14,13*100%/15,7=90% OTVET VIXOD C6H5Br 90%