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Câu 1:
PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Ta có: \(n_{HCl}=0,2\cdot2=0,4\left(mol\right)=n_{NaOH}\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
Câu 2: Bạn xem lại đề !!
\(m_{HCl}=\dfrac{300.7,3\%}{100\%}=21,9g\\ n_{HCl}=\dfrac{21,9}{36,5}=0,6mol\\ HCl+NaOH\rightarrow NaCl+H_2O\left(1\right)\\ n_{NaOH\left(1\right)}=n_{HCl}=0,6mol\\ m_{H_2SO_4}=\dfrac{200.9,8\%}{100\%}=19,6g\\ n_{H_2SO_4}=\dfrac{19,6}{98}=0,2mol\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\left(2\right)\\ n_{NaOH\left(2\right)}=0,2.2=0,4mol\\ n_{NaOH}=0,4+0,6=1mol\\ m_{NaOH}=1.40=40g\\ m_{ddNaOH}=\dfrac{40}{5\%}\cdot100\%=800g\)
\(m_{H_2SO_4}=\dfrac{200.9,8}{100}=19,6\left(g\right)=>n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
PTHH: MO + H2SO4 --> MSO4 + H2O
____0,2<---0,2---------->0,2
=> \(M_{MO}=\dfrac{16}{0,2}=80\left(g/mol\right)\)
=> MM = 64 (g/mol)
=> M là Cu
\(C\%\left(CuSO_4\right)=\dfrac{0,2.160}{16+200}.100\%=14,815\%\)
Bài 1:
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,4\cdot36,5}{14,6\%}=100\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)
Bài 2:
PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{KOH}=\dfrac{100\cdot11,2\%}{56}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{150\cdot9,8\%}{98}=0,15\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,15}{1}\) \(\Rightarrow\) H2SO4 còn dư, KOH p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{K_2SO_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{K_2SO_4}=0,1\cdot174=17,4\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,05\cdot98=4,9\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddKOH}+m_{ddH_2SO_4}=250\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{K_2SO_4}=\dfrac{17,4}{250}\cdot100\%=6,96\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{4,9}{250}\cdot100\%=1,96\%\end{matrix}\right.\)
PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Ta có: \(n_{NaOH}=0,3\cdot0,1=0,03\left(mol\right)=n_{HCl\left(p/ứ\right)}\)
+) Trường hợp 1: NaOH còn dư
PTHH: \(Al+NaOH+H_2O\rightarrow NaAlO_2+\dfrac{3}{2}H_2\uparrow\)
Ta có: \(n_{NaOH\left(dư\right)}=n_{Al}=\dfrac{0,27}{27}=0,01\left(mol\right)\)
\(\Rightarrow n_{HCl}=n_{NaOH\left(p/ứ\right)}=0,02\left(mol\right)\) \(\Rightarrow a=C_{M_{HCl}}=\dfrac{0,02}{0,2}=0,1\left(M\right)\)
+) Trường hợp 2: HCl còn dư
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Ta có: \(n_{Al}=0,01\left(mol\right)\) \(\Rightarrow n_{HCl\left(dư\right)}=0,03\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,06\left(mol\right)\) \(\Rightarrow a=C_{M_{HCl}}=\dfrac{0,06}{0,2}=0,3\left(M\right)\)