\(n_{N_2}=\dfrac{2,24}{22,4}=0,1(mol) \\\Rightarrow n_{Na}=n_{N_2}=0,1(mol)\\ \Rightarrow m_{Na}=0,1.23=2,3(g)\)

\(1,n_{H_2SO_4}=\dfrac{9,8}{98}=0,1(mol)\\ \Rightarrow \text{Số phân tử }H_2SO_4=0,1.6.10^{23}=0,6.10^{23}\\ 2,2Fe(OH)_2+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ \Rightarrow a=2;b=3;c=1;d=6\\ 3,n_{O_2}=\dfrac{320}{32}=10(mol)\)

\(M_X=M_{H_2}.14=2.14=28(g/mol)\)

Chọn B

Câu 10:

Trích mẫu thử, dùng quỳ tím:

- Hóa đỏ: H₂SO₄

- Hóa xanh: NaOH

- Ko đổi màu: Na₂SO₄, BaCl₂

Cho dd H₂SO₄ vào nhóm ko làm quỳ đổi màu:

- Tạo KT trắng: BaCl₂

- Ko hiện tượng: Na₂SO₄

\(BaCl_2+H_2SO_4\to BaSO_4\downarrow+2HCl\)

Câu 11:

\(a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ Al_2O_3+6HCl\to 2AlCl_3+3H_2O\\ \Rightarrow n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ b,n_{Al}=0,1(mol)\Rightarrow m_{Al}=0,1.27=2,7(g)\\ \Rightarrow m_{Al_2O_3}=7,8-2,7=5,1(g)\\ c,n_{Al_2O_3}=\dfrac{5,1}{102}=0,05(mol)\\ \Rightarrow \Sigma n_{HCl}=0,1.3+0,05.6=0,6(mol)\\ \Sigma n_{AlCl_3}=0,1+0,05.2=0,2(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,6.36,5}{10\%}=219(g)\\ \Rightarrow C\%_{AlCl_3}=\dfrac{0,2.133,5}{219+7,8-0,15.2}.100\%=11,79\$\)

\(a,M_A=M_{H_2}.16=2.16=32(g/mol)\\ b,M_A=M_{H_2}.32=2.32=64(g/mol)\\ c,M_A=M_{H_2}.8=2.8=16(g/mol)\)

\(a,\text{Có }0,4.6.10^{23}=2,4.10^{23}(hạt)\\ b,n_{Mg}=\dfrac{120}{24}=5(mol)\\ \Rightarrow \text{Có }5.6.10^{23}=30.10^{23}(hạt)\)

\(a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ \Rightarrow n_{Al}=0,2(mol)\\ b,m_{Al}=0,2.27=5,4(g)\\ \Rightarrow m_{Cu}=11,8-5,4=6,4(g)\)

\(n_{Fe}=\dfrac{1,12}{56}=0,02(ml)\\n_{CuSO_4}=0,2.0,1=0,02(mol)\\ m_{dd_{CuSO_4}}=1,2.200=240(g)\\ a,PTHH:Fe+CuSO_4\to FeSO_4+Cu\\ LTL:\dfrac{0,02}{1}=\dfrac{0,02}{1}\Rightarrow \text{p/ứ hoàn toàn}\\ \Rightarrow n_{FeSO_4}=n_{Cu}=0,02(mol)\\ b,C\%_{FeSO_4}=\dfrac{0,02.152}{1,12+240-0,02.64}.100\%=1,27\%\\ C_{M_{FeSO_4}}=\dfrac{0,02}{0,2}=0,1M\)

\(a,PTHH:CuO+2HCl\to CuCl_2+H_2O\\ b,n_{HCl}=0,2.1=0,2(mol)\\ \Rightarrow n_{CuO}=0,2(mol)\Rightarrow m_{CuO}=0,2.80=16(g)\\ \Rightarrow \%_{CuO}=\dfrac{16}{20,8.}100\%=76,92\%\\ \Rightarrow \%_{Cu}=100\%-76,92\%=23,08\%\)

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ ZnO+2HCl\to ZnCl_2+H_2O\\ b,n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ \Rightarrow n_{Zn}=0,15(mol)\Rightarrow m_{Zn}=0,15.65=9,75(g)\\ \Rightarrow \%_{Zn}=\dfrac{9,75}{25,95}.100\%=37,57\%\\ \Rightarrow \%_{ZnO}=(100-37,57)\%=62,43\%\\ c,n_{ZnO}=\dfrac{25,95-9,75}{81}=0,2(mol)\\ \Rightarrow n_{HCl}=2.0,15+2.0,2=0,7(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,7.36,5}{12\%}=212,92(g)\)