Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
$Fe + 2HCl \to FeCl_2 + H_2$
n H2 = n Fe = 11,2/56 = 0,2(mol)
V H2 = 0,2.22,4 = 4,48(lít)
b)
n HCl = 2n Fe = 0,2.2 = 0,4(mol)
=> CM HCl = 0,4/0,4 = 1M
c)
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
Ta thấy :
n CuO = 64/80 = 0,8 > n H2 = 0,2 nên CuO dư
Theo PTHH :
n CuO pư = n Cu = n H2 = 0,2(mol)
n Cu dư = 0,8 - 0,2 = 0,6(mol)
Vậy :
%m Cu = 0,2.64/(0,2.64 + 0,6.80) .100% = 21,05%
%m CuO = 100% -21,05% = 78,95%
\(n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\\ pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,15 0,15 0,15
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\\
C_M=\dfrac{0,15}{0,1}=1,5M\)
\(n_{Al}=\dfrac{8.1}{27}-0.3\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{49}{98}=0.5\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(2..........3\)
\(0.3..........0.5\)
\(LTL:\dfrac{0.3}{2}< \dfrac{0.5}{3}\Rightarrow H_2SO_4dư\)
\(V_{H_2}=\left(\dfrac{0.3\cdot3}{2}\right)\cdot22.4=10.08\left(l\right)\)
Ta có: \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Xét tỉ lệ: \(\dfrac{0,3}{2}< \dfrac{0,5}{3}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)
Bạn tham khảo nhé!
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Ta có: \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Mg}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b, \(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,3}{3}\), ta được Fe2O3 dư.
Theo PT: \(n_{Fe_2O_3\left(pư\right)}=\dfrac{1}{3}n_{H_2}=0,1\left(mol\right)\Rightarrow n_{Fe_2O_3\left(dư\right)}=0,3-0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe_2O_3}=0,2.160=32\left(g\right)\)
a. 2Al + 6HCl -> 2AlCl3 + 3H2
b. nAl = \(\dfrac{8.1}{27}=0,3\left(mol\right)\)=> \(n_{H_2}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\)
\(V_{H_2}=0,45.22,4=10,08\left(mol\right)\)
a) Pt: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b) nFe = \(\dfrac{11,2}{56}=0,2mol\)
Theo pt: nH2 = nFe = 0,2 mol
=> VH2 = 0,2.22,4 = 4,48lit
c) Theo pt: nHCl = 2nFe = 0,4 mol
=> mHCl = 0,4.36,5 = 14,6 g
=> C% = \(\dfrac{14,6}{73}.100\%=20\%\)
\(2Al(0,3)+6HCl(0,9)--->2AlCl_3(0,3)+3H_2(0,45)\)
\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(m_{HCl}=\dfrac{20.182,5}{100}=36,5\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{36,5}{36,5}=1\left(mol\right)\)
So sánh: \(\dfrac{n_{Al}}{2}=0,15< \dfrac{n_{HCl}}{6}=0,167\)
=> HCl còn dư sau phản ứng, chon nAl để tính.
Khí B là \(H_2:0,45(mol)\)
\(a)\)\(V_{H_2}\left(đktc\right)=10,08\left(l\right)\)
\(b)\)
\(m_{H_2}=0,45.2=0,9\left(g\right)\)
\(mddsau=8,1+182,5-0,9=189,7(g)\)
Dung dich A: \(\left\{{}\begin{matrix}AlCl_3:0,3\left(mol\right)\\HCl\left(dư\right):1-0,9=0,1\left(mol\right)\end{matrix}\right.\)
\(C\%_{AlCl_3}=\dfrac{0,3.133,5}{189,7}.100\%=21,11\%\)
\(C\%_{HCl}\left(dư\right)=\dfrac{0,1.36,5}{189,7}.100\%=1,92\%\)
a) nAl = \(\dfrac{m}{M}\) = \(\dfrac{8,1}{27}=0,3\) (mol)
mHcl p/ư = 182,5 . 20% = 36,5 (g)
\(\Rightarrow n_{HCl}=\dfrac{m}{M}=\dfrac{36,5}{36,5}=1\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
2 (mol) 6 (mol)
0,3(mol) 1(mol)
Lập tỉ lệ và so sánh: \(\dfrac{0,3}{2}< \dfrac{1}{6}\)
Vậy HCl dư
nHidro = \(\dfrac{0,3.3}{2}=0,45\left(mol\right)\)
\(V_{Hidro}=0,45.22,4=10,08\left(l\right)\)