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GIẢI BÀI TẬP BẰNG PHƯƠNG PHÁP TĂNG GIẢM KHỐI LƯỢNG 0 1. Dạng 1: Bài tập về kim loại a. Kim loại + axit (vô cơ hoặc hữu cơ) Ví dụ 1: Hòa tan m gam hỗn hợp A gồm Fe và kim loại M (hóa trị II) trong dung dịch HCl dư thu được 1,008 lít khí (đktc) và dung dịch chứa 4,575 gam muối khan. Giá trị của m là: A. 1,380 B. 1,830 C. 3,180 D. 3,195
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"FK 4 $L!%$" '4,P30?L!/FeFo0 0F#F5 50F5F #0FF# 60F5F Bi gii 9 K 4 kK K D%3 9 ]K , 4 k]K , ]K , F# F $ I 9 #K 4 #K 4 I# # 9 4&'), 4 FK+ ¾¾ ¾ ¾® '4,P30 ?LFe'FoM 1T"Q+"!AL"!/ Fe 4 4 Fe8FK e FFK→ + & Fe ¾¾® & e : ; <&()8 e =948 e <)- @ Fe ¾¾® @ e : ; < '4,=4&'),<9-'- ⇒ @<('4 Fe 4&'), &(H'&4U ('4 ⇒ = = 948e<&(H'&4 e<)4'&4 ⇔ ⇔ ⇒ 50 ;0J "0K&36 >) V dFdR7',X!/!%$",@Z0#Q +"!/%^dR 0FK 4 4 50FK 4 4 #0#!FK 4 9 60FK 4 Bi gii 7',FK 4 ( ¾¾® , 9 K 0?L#?f?!/0 ,FK 4 ( ¾¾® 9 9 K 8,FK 9 8K 9 ,FK 4 ¾¾® 9 9 K : g <,>8H9=9>98&H<9&H ,@FK 4 ¾¾® 9@ 9 K : g <7',=,<-', -', , ('(9- @ 9&H , × ⇒ = = × 4 FK 7', 7,U ('(9- , ⇒ = = × 8H9<7, <49 ⇔ ⇔ #T<9'<H, ⇒ !/%^dR#!FK 4 9 E0J "0K&360BG>(.).5() V dF!&((F 9 #K 4 F#K 4 "%P/$Q %q!%$"H7"GZ0fn;r/$"3"Gn$ 0&Hs),s 50),s&Hs #09Hs.,s 60.,s9Hs Bi gii &(( ( ¾¾® H7F 9 #K 4 ?Lsr/$!/A%n!0 #t"!/F#K 4 AdR ( 4 9 4 9 9 9F#K F #K 8#K 8 K→ 9F#K 4 ¾¾® &F 9 #K 4 : g <9>),=&(H<H9 9@F#K 4 ¾¾® @F 9 #K 4 : g <&((=H7<4& ⇒ @<('- 4 F#K ('- 9 ), ),⇒ = × × = Fe Fo $ F 9 #K 4 F#K 4 9 4 9 4 F #K F #K &(( ), &H &H &((s s &Hs &(( ⇒ = − = × ⇒ = = LL !"#$)+ .).5(+B'.M;C ; )0N= >30O) V duh!P&('H@""A@B""n vD%3b ! 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IK Y ('& 9 ('(- & = = ⇒ Y"+" ⇒ #?f?YJ#KKJx 9 8-', J#KKJx 9 8,'9IKD%3 ¾¾® H'99-J#KKI 9 IK <('( ⇒ Y <('((4 & < s ( ) ( ) 9 9 J #KKJ x 9IK J #KKI 9J xK+ → + &J#KKJx 9 ¾¾® &J#KKI 9 : ; <&HH8J=))89Jx8J<.)=9Jx ('((4 J#KKJx 9 ¾¾® ('((4 J#KKI 9 : ; <(' ⇒ .)=9Jx<9( ⇒ Jx<97 ⇒ Jx# 9 - y QQ !"#$)&()+ )R.GS#@)+ V d#9(4'%N"+"'PP! ^B%%O2" CE ! "#&0#Q" ! "!%$"4&'H)!/0bwBh*""3 ! "#%^ v 0&H 50&(( #049 6049( Bi gii 9( 9 JF 8b* #& ¾¾® 4&'H) 4 JF # ?LbM 1T#?f?!AL"4 9 JF 9 4 JF # JF #+ → & 9 JF ¾¾® & 4 JF # : ; < J 8-9'-= J 8&H<4H'- @ 9 JF ¾¾® @ 4 JF # : ; <4&'H)=9(<&&'H) ⇒ @<('49 9 # JF ('49⇒ = = ⇒ # <('49*B49( 5R.GS#@5)T/ V d?!9'7,z=@ <&,.AaX$D%3 ! " FK0u"Q" ! "Y!g+L!%$"4')9!/50"`T 0 501B@ #0@!" 60 Bi gii 9'7,JF 9 #KK @ 8 FKD%3 ¾¾® 4'),JF 9 #KKF @ 50 z=@' <&,. ?`T"3z=@M <('(90 9 @ 9 @ 9 JF #KK @FK JF #KKF 8@ K+ → & ¾¾® &5: ; <H.@8&H8J=,-@8&H8J<99@ ('(9 ¾¾® ('(95: ; <4')9=9'7,<(')) ⇒ @<9 ⇒ #?f?JF 9 #KK 9 ?%r <&,. ⇒ J8&(H<&,. ⇒ J<,& ⇒ J# 4 - bwB#?#?"3 @!" UU !"#$)G;0 )HG;0 →)+ .).5(+B'. V dK@9'9X %N"+"!%$"4@""A@B"50 #Q+""G!"3 0# 4 # 9 # 9 #K 50# 4 #K #0# 9 <##K 60# 4 # 9 #K Bi gii 9'9J#K ¾¾® 4J#KK5 ?L#?#?0 iKj J#K J#KK→ &J#K ¾¾® &J#KK: ; <J8,-=J897<&H @J#K ¾¾® @J#KK: ; <4=9'9<(') ⇒ @<('(- ⇒ J#K <,,U ⇒ J<&- ⇒ J=# 4 #?#?"3# 4 #K0 5H.('→)G;0 V dK@H"'%N"+"e!%$"-') 0#Q+""G!"3e 0# 4 # 9 K 50# 4 # 9 # 9 K #0# 4 #K# 4 60# 4 K Bi gii H# 98& # 9 Ke ¾¾® -')# 98& #K ?L#?#?"3e0 K@"e!%$" ⇒ "e"Aw" # 98& # 9 K iKj → # 98& #K &# 98& # 9 K ¾¾® &# 98& #K: g <&,849=&,84(<9 @# 98& # 9 K ¾¾® @# 98& #K: g <H=-')<('9 ⇒ @<('& ⇒ e <H(U ⇒ &,849<H( ⇒ <9 "e# 4 # 9 # 9 K VWH9 i&j0##{12"'I|!wR*"g_AwZ"dT"'Fe5uT"}!/" ?0#0 i9j0F!B~I?f$'f$N2g"2"A2T"TR'Fe5uT" }!/"FX0 i4j0?"*6BT"T"0 i,j0?"*T"+ C'Y/9U9((70 i-j05wT"9(&&'$_6{AuT"?0#0