Bài 6:
\(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\\ 2KMnO_4+16HCl_{đặc,nóng}\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\\ 2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ Ta.có:n_{Cl_2}=\dfrac{5}{2}.0,2=0,5\left(mol\right)\\ n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ Vì:\dfrac{0,5}{3}< \dfrac{0,8}{2}\Rightarrow Cl_2thiếu,Aldư\)
Bài 7:
\(n_{MnO_2}=\dfrac{52,2}{87}=0,6\left(mol\right)\\ MnO_2+4HBr_{đặc,nóng}\rightarrow MnBr_2+Br_2+2H_2O\\ 2Fe+3Br_2\rightarrow2FeBr_3\\ n_{Br_2}=n_{MnO_2}=0,6\left(mol\right)\\ n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ Vì:\dfrac{0,3}{2}< \dfrac{0,6}{3}\Rightarrow Br_2dư\)
Bài 8:
\(AgNO_3+HCl\rightarrow AgCl\downarrow\left(trắng\right)+HNO_3\\ n_{AgCl}=\dfrac{4,305}{143,5}=0,03\left(mol\right)\\ n_{HCl}=n_{AgCl}=0,03\left(mol\right)\\ V_{ddHCl}=\dfrac{3}{1,15}=\dfrac{60}{23}\left(ml\right)=\dfrac{3}{1150}\left(l\right)\\ C_{MddHCl}=\dfrac{0,03}{\dfrac{3}{1150}}=11,5\left(M\right)\)
Bài 9:
\(n_{I_2}=\dfrac{76,2}{254}=0,3\left(mol\right)\\ Cl_2+2KI\rightarrow2KCl+I_2\\ n_{KI}=2.n_{I_2}=2.0,3=0,6\left(mol\right)\\ V_{ddKI}=200\left(ml\right)=0,2\left(l\right)\\ \Rightarrow C_{MddKI}=\dfrac{0,6}{0,2}=3\left(M\right)\)
Bài 10:
\(n_{Cl_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ Cl_2+2KBr\rightarrow2KCl+Br_2\\ n_{KBr}=2.n_{Cl_2}=2.0,2=0,4\left(mol\right)\\ m_{ddKBr}=1,34.88,81=119,0054\left(g\right)\\ C\%_{ddKBr}=\dfrac{0,04.119}{119,0054}.100\approx4\%\)
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