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\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\\ V_{HCl}=\dfrac{200}{1000}=0.2L\\ C_M=\dfrac{n_{ct}}{V_{HCl}}=\dfrac{\dfrac{6.5}{65}}{0.2}=0.5mol/l\\ n_{Zn}=\dfrac{m}{M}=\dfrac{6.5}{65}=0.1mol\rightarrow n_{H_2}=0.1mol\rightarrow V_{H_2}=n_{H_2}\cdot22.4=2.24L\)
200ml = 0,2l
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,1.....0,2 0,1 (mol)
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
b) \(V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
PTHH: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
Ta có: \(n_{NaOH}=\dfrac{30\cdot20\%}{40}=0,15\left(mol\right)=n_{CH_3COONa}=n_{CH_3COOH}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CH_3COONa}=0,15\cdot82=12,3\left(g\right)\\C_{M_{CH_3COOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\end{matrix}\right.\)
a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
_____0,2-->0,6----->0,2---->0,3
=> \(C_{M\left(HCl\right)}=\dfrac{0,6}{0,2}=3M\)
b) VH2 = 0,3.24,79 = 7,437 (l)
c) \(C_{M\left(AlCl_3\right)}=\dfrac{0,2}{0,2}=1M\)
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
a, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
c, PT: \(HCl+KOH\rightarrow KCl+H_2O\)
Theo PT: \(n_{KOH}=n_{HCl}=0,4\left(mol\right)\)
\(\Rightarrow m_{ddKOH}=\dfrac{0,4.56}{5,6\%}=400\left(g\right)\)
\(\Rightarrow V_{ddKOH}=\dfrac{400}{1,045}\approx382,78\left(ml\right)\)
\(n_{\left(CH_3COO\right)_2Mg}=\dfrac{0,71}{142}=0,005\left(mol\right)\)
PTHH: 2CH3COOH + Mg ---> (CH3COO)2Mg + H2
0,01<----------------------0,005---------->0,005
\(C_{M\left(CH_3COOH\right)}=\dfrac{0,01}{0,025}=4M\)
PTHH: CH3COOH + NaOH ---> CH3COONa + H2O
0,005--------->0,005
\(V_{ddNaOH}=\dfrac{0,005}{0,75}=\dfrac{1}{150}M\\ V_{H_2}=0,005.22,4=0,0112\left(l\right)\)
\(200ml=0,2l\\ n_{Na_2CO_3}=0,5.0,2=0,1\left(mol\right)\\ PTHH:Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\\ \left(mol\right)........0,1\rightarrow...0,2.......0,2..........0,1.........0,1\\ a,C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\\ b,m_{NaCl}=0,2.58,5=11,7\left(g\right)\\c, V_{ddNaCl}=V_{ddNa_2CO_3}+V_{ddHCl}=0,2+0,2=0,4\left(l\right)\\ C_{M_{NaCl}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)
vì em trộn 2 dung dịch lại với nhau mà, ví dụ em đổ 1 chai nước 500ml vào 1 chai nước 500 ml thì mình phải được 1 lít nước chứ
Gọi \(n_{Fe}=x\left(mol\right)\)\(;n_{Zn}=y\left(mol\right)\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)
Ta có: \(\left\{{}\begin{matrix}56x+65y=18,6\\2x+2y=2n_{H_2}=0,6\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\%m_{Fe}=\dfrac{0,1\cdot56}{18,6}\cdot100\%=30,11\%\)
\(\%m_{Zn}=100\%-30,11\%=69,89\%\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,4
\(n_{HCl}=0,2+0,4=0,6mol\)
\(C_M=\dfrac{n}{V}=\dfrac{0,6}{0,2}=3M\)
\(Fe + 2HCl \rightarrow FeCl_2 + H_2\)
\(n_{Fe}= \dfrac{5,6}{56}= 0,1 mol\)
Theo PTHH:
\(n_{H_2}= n_{Fe}= 0,1 mol\)
Vì ở điều kiện tiêu chuẩn(t độ=25 độ C, áp suất 1Bar), 1 mol khí chiếm thể tích 24,79 lít.
\(\Rightarrow V= 0,1 . 24,79= 2,479 l\)
b)
Theo PTHH:
\(n_{HCl}= 2n_{Fe}= 0,2 mol \Rightarrow m_{HCl}= 0,2 . 36,5=7,3 g\)
\(\Rightarrow m_{dd HCl}= \dfrac{ 7,3 . 100}{10}= 73 g\) (nếu bạn viết ở tử : 7,3 . 100% thì ở mẫu bạn phải viết là 10% nhé)
c) dd sau pư là FeCl2
Ta có: m dd sau pư= mFe + m dung dịch HCl - mH2
= 5,6 + 73 - 0,1 . 2=78,4 g
Theo PTHH:
\(n_{FeCl_2}= n_{Fe}= 0,1 mol \Rightarrow m_{FeCl_2}= 0,1 . 127= 12,7 g\)
C%FeCl2= \(\dfrac{12,7}{78,4}\).100% = 16,199%
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
\(a,n_{NaOH}=0,2.1=0,2\left(mol\right)\\ 2NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ n_{H_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MH_2SO_4}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\ b,n_{Na_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MNa_2SO_4}=\dfrac{0,1}{0,2+0,4}=\dfrac{1}{6}\left(M\right)\\ c,m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
0,2 0,1 ( mol )
\(C_{M_{CH_3COOH}}=\dfrac{0,2}{0,2}=1M\)
2CH3COOH+Mg->(CH3COO)2Mg+H2
0,2---------------------------------------------0,1
n H2=0,1 mol
=>CMCH3COOH=\(\dfrac{0,2}{0,2}=1M\)