\(n_{HCl}=1,5.0,4=0,6(mol)\\ X+HCl\to muối+H_2\)

Bảo toàn H: \(n_{H_2}=\dfrac{n_{HCl}}{2}=0,3(mol)\)

\(\Rightarrow V_{H_2}=0,3.22,4=6,72(l)\\BTKL:m_A+m_{HCl}=m_{muối}+m_{H_2}\\ \Rightarrow m_{muối}=35+0,6.36,5-0,3.2=56,3(g)\)

\(n_{Al}=\dfrac{8,1}{27}=0,3(mol);n_{HCl}=\dfrac{21,9}{36,5}=0,6(mol)\\ a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ LTL:\dfrac{0,3}{2}>\dfrac{0,6}{6}\Rightarrow Al\text { dư}\\ n_{Al(dư)}=0,3-\dfrac{0,6}{3}=0,1(mol)\\ \Rightarrow m_{Al(dư)}=0,1.27=2,7(g)\\ c,n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=0,2(mol)\\n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3(mol)\\ \Rightarrow m_{AlCl_3}=0,2.133,5=26,7(g)\\ d,V_{H_2}=0,3.22,4=6,72(l)\)

PT câu b đâu bạn nhỉ?

\(m_{O_2}=32\left(đvC\right)=32.0,16605.10^{-23}=5,3136.10^{-23}\left(g\right)\)

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\)

Phản ứng thế

\(b,n_{Zn}=\dfrac{1,3}{65}=0,02(mol)\\ \Rightarrow n_{ZnCl_2}=n_{H_2}=0,02(mol)\\ \Rightarrow m_{ZnCl_2}=0,02.136=2,72(g)\\ V_{H_2}=0,02.22,4=0,448(l)\)

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,\text{Số nguyên tử Zn : Số phân tử }HCl : \text{Số phân tử }ZnCl_2 : \text{Số phân tử }H_2=1:2:1:1\\ c,BTKL:m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ \Rightarrow m_{ZnCl_2}=65+70-20=115(g)\)

\(n_{ZnO}=\dfrac{8,1}{81}=0,1(mol)\\ PTHH:2Zn+O_2\xrightarrow{t^o}2ZnO\\ \Rightarrow n_{Zn}=0,1(mol)\\ \Rightarrow m_{Zn}=0,1.65=6,5(g)\)

Câu 1:

\(m_{H_2S}=0,75.34=25,5(g)\\ m_{CaSO_4}=0,025.136=3,4(g)\\ m_{Fe_2O_3}=0,05.160=8(g)\)

Câu 2:

\(V_{N_2}=2,5.22,4=56(l)\\ V_{H_2}=0,03.22,4=0,672(l)\\ V_{O_2}=0,45.22,4=10,08(l)\\ V_{hh}=22,4.(0,2+0,25)=22,4.0,45=10,08(l)\)

Bài 1:

Trong 1 mol B: \(\begin{cases} n_{Na}=\dfrac{106.43,4\%}{23}=2(mol)\\ n_C=\dfrac{106.11,3\%}{12}=1(mol)\\ n_O=\dfrac{106.45,3\%}{16}=3(mol) \end{cases}\)

\(\Rightarrow CTHH_B:Na_2CO_3\)

Câu 2:

\(a,4K+O_2\xrightarrow{t^o}2K_2O\\ 4:1:2\\ b,2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3\\ 2:3:2\\ c,2Al+6HCl\to 2AlCl_3+3H_2\\ 2:6:2:3\\ d,Na_2O+H_2O\to 2NaOH\\ 1:1:2\\ \)

\(e,2KClO_3\xrightarrow{t^o}2KCl+3O_2\\ 2:2:3\\ f,Zn+2HCl\to ZnCl_2+H_2\\ 1:2:1:1\\ g,2Fe(OH)_3\xrightarrow{t^o}Fe_2O_3+3H_2O\\ 2:1:3\\ h,P_2O_5+3H_2O\to 2H_3PO_4\\ 1:3:2\)