\(a,M_{R_2O}=\dfrac{18,6}{0,3}=62(g/mol)\\ \Rightarrow M_R=\dfrac{62-16}{2}=23(g/mol)\\ \Rightarrow R:Na\\ b,CTHH:RO_3\\ \Rightarrow M_{RO_3}=\dfrac{16}{0,2}=80(g/mol)\\ \Rightarrow M_R+4=80\\ \Rightarrow M_R=32(g/mol)\ \Rightarrow R:S\)

\(CTHH_A:N_xO_y\\ M_A=23.2=46(g/mol)\\ \Rightarrow 14x+16y=46\)

Với \(x=1\Rightarrow y=2(nhận)\)

\(\Rightarrow CTHH_A:NO_2\)

\(a,m_{H_2S}=0,4.34=13,6(g);V_{H_2S}=0,4.22,4=8,96(l)\\ m_{SO_2}=0,025.64=1,6(g);V_{SO_2}=0,025.22,4=0,56(l)\\ m_{NO}=0,22.30=6,6(g);V_{NO}=0,22.22,4=4,928(l)\)

\(b,m_{CO}=0,45.28=12,6(g);V_{H_2S}=0,45.22,4=10,08(l)\\ m_{NH_3}=0,45.17=7,65(g);V_{NH_3}=0,45.22,4=10,08(l)\\ m_{CH_4}=0,45.16=7,2(g);V_{CH_4}=0,45.22,4=10,08(l)\\ m_{CO_2}=0,45.44=19,8(g);V_{CO_2}=0,45.22,4=10,08(l)\)

\(c,m_{hh}=0,1.28+0,3.48+0,375.36,5=30,8875(g)\\ V_{hh}=22,4.(0,1+0,3+0,375_17,36(l)\\ d,n_{O_2}=\dfrac{6.10^{23}}{6.10^{23}}=1(mol)\\ \Rightarrow m_{O_2}=32(g);V_{O_2}=22,4(l)\\ n_{N_2O_5}=\dfrac{7,2.10^{23}}{6.10^{23}}=1,2(mol)\\ \Rightarrow m_{N_2O_5}=1,2.108=129,6(g);V_{N_2O_5}=26,88(l)\\ n_{CO}=\dfrac{4,5.10^{23}}{6.10^{23}}=0,75(mol)\\ \Rightarrow m_{CO}=0,75.28=21(g);V_{CO}=0,75.22,4=16,8(l)\)

Câu 3:

\(a,PTHH:Mg+2HCl\to MgCl_2+H_2\\ b,n_{Mg}=\dfrac{4,8}{24}=0,2(mol)\\ \Rightarrow n_{MgCl_2}=0,2(mol)\\ \Rightarrow m_{MgCl_2}=0,2.95=19(g)\\ c,n_{H_2}=0,2(mol)\Rightarrow V_{H_2}=0,2.22,4=4,48(l)\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ a,2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,1(mol);n_{HCl}=0,3(mol)\\ b,m_{Al}=0,1.27=2,7(g);m_{HCl}=0,3.36,5=10,95(g)\\ m_{AlCl_3}=0,1.133,5=13,35(g)\\ c,n_{Al}=\dfrac{16,2}{27}=0,6(mol)\\ \Rightarrow n_{H_2}=1,5n_{Al}=0,9(mol)\\ \Rightarrow V_{H_2}=0,9.22,4=20,16(l)\)

\(V_{CH_4(nguyên chất)}=1000.98\%=980(l)\\ \Rightarrow n_{CH_4}=\dfrac{980}{22,4}=43,75(mol)\\ CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O\\ \Rightarrow n_{O_2}=87,5(mol)\\ \Rightarrow V_{O_2}=87,5.22,4=1960(l)=1,96(m^3)\)

Câu 1:

Đặt \(n_{Al}=x(mol);n_{Fe}=y(mol)\Rightarrow 27x+56y=0,83(1)\)

\(n_{H_2}=\dfrac{0,56}{22,4}=0,025(mol)\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow 1,5x+y=0,025(2)\\ (1)(2)\Rightarrow x=y=0,01(mol)\\ \Rightarrow \%_{Al}=\dfrac{0,01.27}{0,83}.100\%=32,53\%\\ \Rightarrow \%_{Fe}=100\%-32,53\%=67,47\%\)

Câu 2:

Đặt \(n_{Al}=x(mol);n_{Mg}=y(mol)\Rightarrow 27x+24y=4,5(1)\)

\(n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Mg+H_2SO_4\to MgSO_4+H_2\\ \Rightarrow 1,5x+y=0,225(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,075(mol)\\ \Rightarrow \%_{Al}=\dfrac{0,1.27}{4,5}.100\%=60\%\\ \Rightarrow \%_{Mg}=100\%-60\%=40\%\)

Dạng PP hai dòng:

\(PTHH:2A+Cl_2\to 2ACl\\ \Rightarrow n_A=n_{ACl}\\ \Rightarrow \dfrac{9,2}{M_A}=\dfrac{23,4}{M_A+35,5}\\ \Rightarrow M_A=23(g/mol)\)

Vậy A là natri

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ a,Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{Zn}=0,15(ml);n_{HCl}=0,3(mol)\\ a,m_{Zn}=0,15.65=9,75(g)\\ b,C_{M_{HCl}}=\dfrac{0,3}{0,1}=3M\)

\(M_{H_xP_yO_z}=M_{H_xPO_z}=\dfrac{31}{31,63\%}=98(g/mol)\)

\(a,=5x^2-5x+3x-3=\left(x-1\right)\left(5x+3\right)\\ b,=2x^2-5x+2x-5=\left(2x-5\right)\left(x+1\right)\\ c,=x^2+5x-3x-15=\left(x+5\right)\left(x-3\right)\\ d,=7x^2-7x+x-1=\left(x-1\right)\left(7x+1\right)\)