a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

b, \(n_{O_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

Theo PT: \(n_{Al_2O_3}=\dfrac{2}{3}n_{O_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)

c, \(H=\dfrac{18,36}{20,4}.100\%=90\%\)

\(\veebar\)

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)

\(\Rightarrow m=m_{Cu}=12-5,6=6,4\left(g\right)\)

b, \(n_{FeSO_4}=n_{H_2SO_4\left(pư\right)}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,2.2-0,1=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{FeSO_4}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\\C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\end{matrix}\right.\)

\(a)n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(\dfrac{1}{6}\)           \(0,5\)             \(\dfrac{1}{6}\)              \(0,25\)

\(\%m_{Al}=\dfrac{1:6.27}{25}\cdot100=18\%\\ \%m_{Al_2O_3}=100-18=82\%\\ b)n_{Al_2O_3}=\dfrac{25-1:6.27}{102}=\dfrac{41}{204}mol\\ Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

\(\dfrac{41}{204}\)           \(\dfrac{41}{36}\)            \(\dfrac{41}{102}\)

\(m_{ddHCl}=\dfrac{\left(0,5+41:36\right)36,5}{20}\cdot100=299,1g\\ V_{ddHCl}=\dfrac{299,1}{1,1}=271,9ml\)

\(m_{dd}=299,1+25-0,25.2=323,6g\)

\(m_{AlCl_3}=\left(\dfrac{1}{6}+\dfrac{41}{102}\right)\cdot133,5=75,9g\\ C_{\%AlCl_3}=\dfrac{75,9}{323,6}\cdot100=23,45\%\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

a, \(n_{H_2}=\dfrac{0,7437}{24,79}=0,03\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{H_2}=0,06\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{0,06}{0,1}=0,6\left(l\right)\)

b, \(H=\dfrac{669,33}{743,7}.100\%=90\%\)

Câu 1: Gọi kim loại là M

\(M+H_2SO_4\rightarrow MSO_4+H_2\)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Theo pt: \(n_M=n_{H_2}=0,05mol\)

\(M_M=\dfrac{1,2}{0,05}=24\)

Vậy kim loại là Mg, đáp án A

Câu 2: Chọn A, Cu đứng sau H trong dãy hoạt động hóa học nên sẽ không tác dụng với các axit trên

1,c

2,d

a,Mg+2HCl=>MgCl2+H2

b,nHCl=0,05.3=0,15(mol)

nMg=12/24=0,5(mol)=>Mg dư, tính thao HCl

nH2=1/2 nHCl=0,075(mol)

=>VH2=0,075.22,4=1,68(l)

c,nMgCl2=nH2=0,075(mol)

mMgCl2=0,075.95=7,125(g)

a)\(Mg+2HCl\rightarrow MgCl_2+H_2\)

b) \(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)

\(n_{HCl}=0,05.3=0,15\)

Ta có \(\dfrac{0,15}{2}< \dfrac{0,5}{1}\)nên Mg dư, tính theo HCl

\(n_{H_2}=\dfrac{n_{HCl}}{2}=0,075\left(mol\right)\)

\(V_{H_2}=0,075.22,4=1,68\left(l\right)\)

c) \(n_{MgCl_2}=\dfrac{n_{HCl}}{2}=0,075\left(mol\right)\)

\(m_{MgCl_2}=0,075.95=7,125g\)

Câu 1:

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right);n_{HCl}=0,3.1=0,3\left(mol\right)\)

Ta có \(\dfrac{0,2}{1}>\dfrac{0,3}{2}\Rightarrow\) Fe dư, vậy khối lượng muối tính theo HCl

Theo pt: \(n_{FeCl_2}=\dfrac{n_{HCl}}{2}=0,15mol\)

\(m_{FeCl_2}=0,15.127=19,05g\)

Chọn đáp án D

Câu 2

\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

\(n_{NaOH}=0,2.2=0,4mol\)

Theo pt: \(n_{H_2SO_4}=\dfrac{n_{NaOH}}{2}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2SO_4}=\dfrac{0,2}{0,5}=0,4\left(l\right)=400ml\)

Chọn đáp án B

\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ n_{Cu\left(LT\right)}=n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\\ n_{Cu\left(TT\right)}=\dfrac{2}{64}=0,03125\left(mol\right)\\ \Rightarrow H=\dfrac{0,03125}{0,05}.100\%=62,5\%\)