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\(n_{Na}=\dfrac{6,9}{23}=0,3\left(mol\right)\\ m_{HCl}=200.3,65\%=7,3\left(g\right)\\ n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)
\(PTHH:2Na+2HCl\rightarrow2NaCl+H_2\uparrow\\ LTL:0,3>0,2\Rightarrow Na.dư\)
Theo pt: nH2 = 2nHCl = 2.0,2 = 0,4 (mol)
VH2 = 0,4.22,4 = 8,96 (l)
Theo pt: nNaCl = nNa (phản ứng) = nHCl = 0,2 (mol)
=> \(\left\{{}\begin{matrix}m_{NaCl}=0,2.58,5=11,7\left(g\right)\\m_{Na\left(dư\right)}=\left(0,3-0,2\right).23=2,3\left(g\right)\\m_{H_2}=0,4.2=0,8\left(g\right)\end{matrix}\right.\)
=> \(m_{dd}=200+6,9-2,3-0,8=203,8\left(g\right)\)
=> C%NaCl = \(\dfrac{11,7}{203,8}=5,74\%\)
\(n_{Al} = a\ ; n_{Fe} =b\\ \Rightarrow 27a + 56b = 11(1)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = 1,5a + b = \dfrac{8,96}{22,4} = 0,4(2)\\ (1)(2) \Rightarrow a = 0,2 ; b = 0,1\\ n_{HCl\ dư} = \dfrac{200.21,9\%}{36,5} - 0,2.3 - 0,1.2 = 0,4(mol)\\ m_{dd\ sau\ pư} = 11 + 200 - 0,4.2 = 210,2(gam)\\ C\%_{HCl} = \dfrac{0,4.36,5}{210,2}.100\% = 6,95\%\\ \)
\(C\%_{AlCl_3} = \dfrac{0,2.133,5}{210,2}.100\% = 12,7\%\\ C\%_{FeCl_2} = \dfrac{0,1.127}{210,2}.100\% = 6,04\%\)
Bài 1:
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
Ta có: \(n_{Mg}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\cdot0,11\cdot1,5=0,0825\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,0825\cdot24=1,98\left(g\right)\)
Ta có: \(n_{SO_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)
\(m_{H_2SO_4}=300.78,4\%=235,2\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{235,2}{98}=2,4\left(mol\right)\)
PT: \(2Fe+6H_2SO_{4\left(đ\right)}\underrightarrow{t^o}Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)
____0,5____1,5________0,25______0,75 (mol)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=2,4-1,5=0,9\left(mol\right)\)
Ta có: m dd sau pư = mFe + m dd H2SO4 - mSO2
= 0,5.56 + 300 - 0,75.64 = 280 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,9.98}{280}.100\%=31,5\%\\C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,25.400}{280}.100\%\approx35,7\%\end{matrix}\right.\)
Bạn tham khảo nhé!
\(2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2\)
\(n_{Al}= \dfrac{10,8}{27}=0,4 mol\)
\(n_{HCl}= 0,3 mol\)
\(2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2\)
Trước PƯ: 0,4 0,3
PƯ: 0,1 0,3 0,1 0,15
Sau PƯ: 0,3 0 0,1 0,15
Đây là mình làm tắt, bạn nên làm bài hết dư này theo cách của bạn
Sau PƯ: dd X có AlCl3 (0,1 mol) và 0,15 mol H2
\(V= 0,15 . 22,4=3,36 l\)
\(m_{dd sau pư}= m_{Al pư} + m_{dd HCl} - m_{H_2}= 0,1 . 27 + 300 - 0,15. 2=302 , 4g\)
C% X=\(\dfrac{ 0,1 . 133,5}{302,4}\). 100%= 4,41%
\(n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\\ C_{M\left(HCl\right)}=\dfrac{0,1}{0,2}=0,5M\)
a) Ta có: \(m_{H_2SO_4}=120\cdot10\%=12\left(g\right)\) \(\Rightarrow C\%_{H_2SO_4}=\dfrac{12}{120+30}\cdot100\%=8\%\)
b) Ta có: \(m_{KOH}=195\cdot8\%=15,6\left(g\right)\) \(\Rightarrow C\%_{KOH}=\dfrac{15,6+5}{195+5}\cdot100\%=10,3\%\)
1.
\(\%C=\dfrac{12}{44}.100\simeq22,73\%\)
2.
\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ V=\dfrac{n}{C_M}=\dfrac{0,1}{2}=0,05\left(M\right)\)
\(C_{M\left(FeCl_2\right)}=\dfrac{n}{V}=\dfrac{0,05}{0,05}=1\left(M\right)\)
a)SO3+H2O---->H2SO4
\(n_{SO3}=\frac{20}{80}=0,25\left(mol\right)\)
\(n_{H2SO4}=n_{SO3}=0,25\left(mol\right)\)
\(m_{H2SO4}=0,25.98=24,5\left(g\right)\)
\(m_{dd}=20+180=200\left(g\right)\)
\(C\%=\frac{24,5}{200}.100\%=12,25\%\)
b) \(Fe+H2SO4-->FeSO4+H2\)
\(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\)
\(\Rightarrow H2SO4dư\)
dd B gồm H2SO4 dư và FeSO4
\(m_{H2}=0,4\left(g\right)\)
\(m_{ddB}=m_{ddH2SO4}+m_{Fe}-m_{H2}=200+11,2-0,4=210,8\left(g\right)\)
\(n_{H2SO4}=n_{FE}=0,2\left(mol\right)\)
\(n_{H2SO4}dư=0,25-0,2=0,05\left(mol\right)\)
\(m_{H2SO4}dư=0,05.98=4,9\left(g\right)\)
\(C\%_{H2SO4}=\frac{4,9}{210,8}.100\%=2,32\%\)
\(n_{FeSO4}=n_{Fe}=0,2\left(mol\right)\)
\(m_{FeSO4}=0,2.152=30,4\left(g\right)\)
\(C\%_{FeSO4}=\frac{30,4}{210,8}.100\%=14,42\%\)
Sao mH2=0,4(g)