Đáp án: hoà tan 10,6g hỗn hợp gồm Znvà ZnO bằng dung dịch HCl 7,3% vừa đủ Sau khi phản ứng kết thúc, thu được 2,24l khí H2 ở (dktc) a) Tính % khối lượng mỗi chất trong hỗn hợp ban đầu? b) Tính khối lượng muối thu được sau phản ứng c) Tính khối lượng dung dịch HCl 7,3% cần... - Đáp án a) %Zn= 61,32% %ZnO= 38,68% b) mZnCl2= 20,4 g c) mdd HCl= 150 g Giải thích các bước giải a)nH2= 0,1 molPTHHZn + 2HCl→ ZnCl2+ H2 0,1 0,2 0,1 0,1ZnO+ 2HCl→ZnCl2+H2O⇒ nZnO= ( 10,6-0,156)/81= 0,05 mol%Zn= 6,5/10,6100%= 61,32%%ZnO= 38,68%b) mZnCl2= (0,1+0,05)136=20,4 gc)mdd HCl= (0,15236,5)/7,3%=1...

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2 năm trước

hoà tan 10,6g hỗn hợp gồm Znvà ZnO bằng dung dịch HCl 7,3% vừa đủ Sau khi phản ứng kết thúc, thu được 2,24l khí H2 ở (dktc). a) Tính % khối lượng mỗi chất trong hỗn hợp ban đầu? b) Tính khối lượng muối thu được sau phản ứng c) Tính khối lượng dung dịch HCl 7,3% cần dùng

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Trang Luong Nguyen

2 năm trước

Đáp án:

%Zn= 61,32%

%ZnO= 38,68%

mZnCl2= 20,4 g

mdd HCl= 150 g

Giải thích các bước giải:

a)
nH2= 0,1 mol
PTHH:
Zn + 2HCl→ ZnCl2+ H2
0,1. 0,2. 0,1. 0,1
ZnO+ 2HCl→ZnCl2+H2O
⇒ nZnO= ( 10,6-0,1.56)/81= 0,05 mol
%Zn= 6,5/10,6.100%= 61,32%
%ZnO= 38,68%

b)
mZnCl2= (0,1+0,05).136=20,4 g
c)
mdd HCl= (0,15.2.36,5)/7,3%=150 g

nguyentransondhy

2 năm trước

Đáp án:

a) %mZn=61,32% -> %mZnO=38,68%

b) mZnCl2=20,4 gam

c) m dd HCl=150 gam

Giải thích các bước giải:

a)Zn + 2HCl ->ZnCl2 + H2

ZnO + 2HCl -> ZnCl2 + H2O

ta có: nH2=2,24/22,4=0,1 mol =nZn (theo ptpu)

-> mZn=0,1.65=6,5 gam -> mZnO=10,6-6,5=4,1 gam -> nZnO=4,1/(65+16)=0,05 mol

%mZn=6,5/10,6=61,32% -> %mZnO=38,68%

b) Muối là ZnCl2 -> nZnCl2=nZn +nZnO=0,1+0,05=0,15 mol -> mZnCl2=0,15.(65+35,5.2)=20,4 gam

c) Theo ptpu: nHCl=2nZnCl2=0,3 mol -> mHCl=0,3.36,5=10,95 gam

-> m dung dịch HCl=10,95/7,3%=150 gam

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