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a)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,15-->0,3----->0,15-->0,15
=> \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
b) \(m_{FeCl_2}=0,15.127=19,05\left(g\right)\)
c) \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
=> \(C\%=\dfrac{10,95}{200}.100\%=5,475\%\)
Bài 1
\(a,n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ CuO+2HCl\xrightarrow[]{}CuCl_2+H_2O\\ n_{CuCl_2}=n_{CuO}=0,2mol\\ m_{CuCl_2}=0,2.135=27\left(g\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ C_{MHCl}=\dfrac{0,4}{0,5}=0,8\left(M\right)\)
Bài 5
\(a,n_{NaOH}=0,2.1=0,2\left(mol\right)\\ 2NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ n_{H_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MH_2SO_4}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\ b,n_{Na_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MNa_2SO_4}=\dfrac{0,1}{0,2+0,4}=\dfrac{1}{6}\left(M\right)\\ c,m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
_____0,1-->0,2----->0,1----->0,1
=> VH2 = 0,1.22,4 = 2,24 (l)
b) \(V_{ddHCl}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)
c) \(C_{M\left(ZnCl_2\right)}=\dfrac{0,1}{0,5}=0,2M\)
Tham khảo nha :
Link:Câu hỏi của Nguyễn Nhật Tiên Tiên - Hóa học lớp 8 | Học trực tuyến
Al2O3 + 6HCl -> 2AlCl3 + 3H2O (1)
nAl2O3=0,1(mol)
nHCl=1(mol)
Từ 1:
nHCl=6nAl2O3=0,6(mol)<1 nên sau PƯ còn 0,4 mol HCl dư
nAlCl3=2nAl2O3=0,2(mol)
mAlCl3=133,5.0,2=26,7(g)
mHCl dư=0,4.36,5=14,6(g)
C% dd AlCl3=\(\dfrac{26,7}{400+10,2}.100\%=6,51\%\)
C% dd HCl dư=\(\dfrac{14,6}{410,2}.100\%=3,56\%\)
a) \(n_{Fe}=\dfrac{1,12}{56}=0,02\left(mol\right)\)
PTHH: Fe + 2HCl -->FeCl2 + H2
_____0,02->0,04--->0,02--->0,02
=> VH2 = 0,02.22,4 = 0,448(l)
b) mFeCl2 = 0,02.127 = 2,54(g)
c) \(C_{M\left(HCl\right)}=\dfrac{0,04}{0,2}=0,2M\)
Fe + 2HCl → FeCl2 + H2
1 2 1 1
0,02 0,04 0,02 0,02
nFe=\(\dfrac{1,12}{56}\)= 0,02(mol)
a). nH2=\(\dfrac{0,02.1}{1}\)= 0,02(mol)
→VH2= n . 22,4 = 0,02 . 22,4 = 0,448(l)
b). nFeCl2= \(\dfrac{0,02.1}{1}\)= 0,02(mol)
→mFeCl2= n . M = 0,02 . 127 = 2,54(g)
c). 200ml = 0,2l
nHCl= \(\dfrac{0,02.2}{1}\)=0,04(mol)
→CM= \(\dfrac{n}{V}\)= \(\dfrac{0,04}{0,2}\)= 0,2M
a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
b, Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Mg}=0,8\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)
c, Theo PT: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)
Al2O3 + 6HCl → 2AlCl3 + 3H2O
\(n_{Al_2O_3}=\frac{10,2}{102}=0,1\left(mol\right)\)
a) Theo PT: \(n_{HCl}=6n_{AlCl_3}=6\times0,1=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,6\times36,5=21,9\left(g\right)\)
\(\Rightarrow C\%_{HCl}=\frac{21,9}{400}\times100\%=5,475\%\)
b) \(m_{dd}saupư=10,2+400=410,2\left(g\right)\)
c) Theo pT: \(n_{AlCl_3}=2n_{Al_2O_3}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,2\times133,5=26,7\left(g\right)\)
\(\Rightarrow C\%_{AlCl_3}=\frac{26,7}{410,2}\times100\%=6,51\%\)
nAl2O3= 10.2/102=0.1 mol
Al2O3 + 6HCl --> 2AlCl3 + 3H2O
0.1______0.6______0.2
mHCl = 0.6*36.5=21.9g
C%HCl = 21.9*100/400=5.475%
mdd sau phản ứng = 10.2+400=410.2 g
mAlCl3 = 0.2*133.5=26.7 g
C%AlCl3 = 26.7/410.2*100%= 6.51%