Tính khối lượng dd HCl 7,3% đủ để trung hòa hết 200ml dd NaOH 0,1M?
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Bài 13 :
\(a)n_{Fe_2O_3} = \dfrac{9,6}{160} = 0,06(mol)\\ Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O\\ n_{HCl} = 6n_{Fe_2O_3} = 0,36(mol)\\ C\%_{HCl} = \dfrac{0,36.36,5}{150}.100\% = 8,76\%\\ \Rightarrow X = 8,76 b) n_{FeCl_3} = 2n_{Fe_2O_3} = 0,12(mol)\\ m_{FeCl_3} = 0,12.162,5 =19,5(gam)\)
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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
\(a,n_{NaOH}=0,2.1=0,2\left(mol\right)\\ 2NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ n_{H_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MH_2SO_4}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\ b,n_{Na_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MNa_2SO_4}=\dfrac{0,1}{0,2+0,4}=\dfrac{1}{6}\left(M\right)\\ c,m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)
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1)
a,\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
\(C_{M_{ddNaOH}}=\dfrac{0,2}{0,242}=0,83M\)
\(C\%_{ddNaOH}=\dfrac{8.100\%}{242}=3,3\%\)
b,\(n_{H_2SO_4}=0,1.0,15=0,015\left(mol\right)\)
PTHH: 2NaOH + H2SO4 → Na2SO4 + 2H2O
Mol: 0,03 0,015
\(C_{M_{ddNaOH}}=\dfrac{0,03}{0,2}=0,15M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,1 0,2
a) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(C_{M_{ddHCl}}=\dfrac{0,2}{1,5}=0,13\left(M\right)\)
b) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)
1 1 1 1
0,2 0,2
\(n_{NaOH}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(m_{NaOH}=0,2.40=8\left(g\right)\)
\(m_{ddNaOH}=\dfrac{8.100}{5}=160\left(g\right)\)
Chúc bạn học tốt
\(a,n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{Mg}=0,2\left(mol\right)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,2}{1,5}=\dfrac{2}{15}M\\ b,n_{HCl}=\dfrac{2}{15}\cdot0,75=0,1\left(mol\right)\\ PTHH:HCl+NaOH\rightarrow NaCl+H_2O\\ \Rightarrow n_{NaOH}=n_{HCl}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{NaOH}}=0,1\cdot40=4\left(g\right)\\ \Rightarrow m_{dd_{NaOH}}=\dfrac{4\cdot100\%}{5\%}=80\left(g\right)\)
\(NaOH + HCl \to NaCl + H_2O\\ n_{HCl} = n_{NaOH} = 0,2.0,1 = 0,02(mol)\\ \Rightarrow m_{dd\ HCl} = \dfrac{0,02.36,5}{7,3\%} = 10\ gam \)