\(4K+O_2\rightarrow\left(t^o\right)2K_2O\\ 1.n_K=\dfrac{5,85}{39}=0,15\left(mol\right)\\ \Rightarrow n_{K_2O}=\dfrac{0,15}{2}=0,075\left(mol\right)\\ m_{K_2O}=94.0,075=7,05\left(g\right)\\ 2,n_K=\dfrac{9,36}{39}=0,24\left(mol\right)\\ n_{O_2}=\dfrac{0,24}{4}=0,06\left(mol\right)\\ V_{O_2\left(đkc\right)}=0,06.24,79=1,4874\left(l\right)\)

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\(Câu.2:\\ 2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ n_{KClO_3}=\dfrac{14,7}{122,5}=0,12\left(mol\right)\\ n_{KCl}=n_{KClO_3}=0,12\left(mol\right);n_{O_2}=\dfrac{3}{2}.0,12=0,18\left(mol\right)\\ V_{O_2\left(đkc\right)}=0,18.24,79=4,4622\left(l\right)\\ m_{KCl}=74,5.0,12=8,94\left(g\right)\)

Câu 3:

\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\\ 1,V_{H_2\left(đkc\right)}=24,79.0,45=11,1555\left(l\right)\\ 2,n_{HCl}=\dfrac{6}{2}.0,3=0,9\left(mol\right)\\ V_{ddHCl}=\dfrac{0,9}{1,5}=0,6\left(l\right)\\ 3,n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\\ V_{ddsau}=V_{ddHCl}=0,6\left(l\right)\\ C_{MddAlCl_3}=\dfrac{0,3}{0,6}=0,5\left(M\right)\)

- Nén khí CO₂ ở áp suất cao giúp tăng độ tan của khí trong nước ngọt.

\(n_{CuSO_4}=\dfrac{10}{160}=0,0625\left(mol\right)\\ C_{MddCuSO_4}=\dfrac{0,0625}{0,2}=0,3125\left(M\right)\\ m_{ddCuSO_4}=200.1,26=252\left(g\right)\\ C\%_{ddCuSO_4}=\dfrac{10}{252}.100\%\approx3,968\%\)

Hợp chất có công thức: \(MgO\)

PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

\(n_{Mg}=\dfrac{7,2}{24}=0,3mol\) \(\Rightarrow n_{MgO}=0,3mol\)

\(m_{MgO}=0,3\cdot\left(24+16\right)=12g\)

\(2Mg+O_2\rightarrow\left(t^o\right)MgO\\ n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\\ n_{MgO}=n_{Mg}=0,3\left(mol\right)\\ m_{hc}=m_{MgO}=0,3.40=12\left(g\right)\)

a) 250ml=0,25L

\(n_{FeSO_4\left(1,2M\right)}=v\cdot C_M=0,25\cdot1,2=0,3\left(mol\right)\)

\(n_{FeSO_4\left(1,5M\right)}=v\cdot C_M=0,25\cdot1,5=0,375\left(mol\right)\)

b) \(n_{FeSO_4}=n_{FeSO_4\left(1,5M\right)}-n_{FeSO_4\left(1,2M\right)}=0,375-0,3=0,075\left(mol\right)\)

\(m_{FeSO_4}=n\cdot M=0,075\cdot152=11,4\left(g\right)\)

\(a)n_{FeSO_4\left(2\right)}=0,25.1,2=0,3mol\\ n_{FeSO_4\left(sau\right)}=0,25.1,5=0,375mol\\ b.n_{FeSO_4\left(1\right)}=0,375-0,3=0,075mol\\ m=m_{FeSO_4\left(1\right)}=0,075.152=11,4g\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

Theo PT: \(n_{Fe_3O_4\left(LT\right)}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{30}\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4\left(LT\right)}=\dfrac{1}{30}.232=\dfrac{116}{15}\left(g\right)\)

\(\Rightarrow H=\dfrac{\dfrac{116}{15}}{11,6}.100\%\approx66,67\%\)

\(n_{Fe_3O_4}=\dfrac{1}{30}\left(mol\right)\) chứ không phải \(\dfrac{1}{3}\) em nhé.

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo pt: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2mol\)

\(n_{Fe}=\dfrac{6,72}{56}=0,12mol\\ 2Fe+O_2\xrightarrow[]{t^0}2FeO\)

0,12        0,06       0,12

\(m_{FeO}=0,12.72=8,64g\\ V_{O_2}=0,06.24,79=1,4874l\)

fgsr

4Fe\(_{ }\) + 3O2 -> 2Fe2O3