Bài 9:

\(n_{Fe}=\dfrac{2,8}{56}=0,2\left(mol\right)\)

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

0,2-->0,4---->0,2------>0,2

a. \(V_{HCl}=\dfrac{0,4}{1}=0,4\left(l\right)=400\left(ml\right)\)

b.\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c. \(CM_{FeCl_2}=\dfrac{0,2}{0,4}=0,5M\)

Bài 10:

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

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

0,1<--0,2<----0,1<----0,1

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

b. \(V_{HCl}=\dfrac{0,2}{0,1}=2\left(l\right)=200\left(ml\right)\)

c.\(CM_{FeCl_2}=\dfrac{0,1}{2}=0,05M\)

Bài 8:

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,1--->0,15------>0,05---------->0,15

a. \(V_{H_2SO_4}=\dfrac{0,15}{1}=0,15\left(l\right)=150\left(ml\right)\)

b. \(V'=0,15.22,4=3,36\left(l\right)\)

c. \(CM_{Al_2\left(SO_4\right)_3}=\dfrac{0,05}{0,15}=\dfrac{1}{3}M\)

22

a

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

x--------------->x--------->x

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

y-------------->y--------->y

Có hệ: \(\left\{{}\begin{matrix}65x+56y=12,1\\136x+127y=26,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

\(\%m_{Zn}=\dfrac{0,1.65.100\%}{12,1}=53,72\%\\ \%m_{Fe}=\dfrac{0,1.56.100\%}{12,1}=46,28\%\)

b

\(V_{H_2}=\left(0,1+0,1\right).22,4=4,48\left(l\right)\)

23

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

x----------------------------->x

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

y-------------------------------->1,5y

Có hệ: \(\left\{{}\begin{matrix}24x+27y=3,75\\x+1,5y=\dfrac{3,92}{22,4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

\(\%m_{Mg}=\dfrac{0,1.24.100\%}{3,75}=64\%\\ \%m_{Al}=\dfrac{0,05.27.100\%}{3,75}=36\%\)

\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

a)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

x------->2x

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

y--------->6y

Có hệ: \(\left\{{}\begin{matrix}2x+6y=0,5\\80x+160y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)

\(m_{CuO}=0,1.80=8\left(g\right)\\ m_{Fe_2O_3}=0,05.160=8\left(g\right)\)

b

\(\%m_{CuO}=\dfrac{0,1.80.100\%}{16}=50\%\\ \%m_{Fe_2O_3}=\dfrac{0,05.160.100\%}{16}=50\%\)

Gọi \(n_{CO}=x\left(mol\right),n_{CO_2}=y\left(mol\right)\)

\(Có:28x+44y=0,3\left(1\right)\)

Mặt khác:

\(\dfrac{m_{CO}+m_{CO_2}}{n_{CO}+n_{CO_2}}=15.2\\ \Leftrightarrow\dfrac{28x+44y}{x+y}=30\)

\(\Leftrightarrow x+y=\dfrac{0,3}{300}=\dfrac{1}{100}\left(mol\right)\)

Vậy số mol của hỗn hợp khí là \(\dfrac{1}{100}\left(mol\right)\)

Trong 300g dung dịch \(Ba\left(OH\right)_2\) 10% có:

\(m_{Ba\left(OH\right)_2}=\dfrac{300.10\%}{100\%}=30\left(g\right)\)

Khối lượng dung dịch \(Ba\left(OH\right)_2\) 25% cần pha:

\(m_{dd.Ba\left(OH\right)_2}=\dfrac{30.100\%}{25\%}=120\left(g\right)\)

\(m_{nước.cần.dùng}=300-120=180\left(g\right)\)

Cách pha:

Cần 120 gam dung dịch \(Ba\left(OH\right)_2\) 25% và 180 gam nước cất (hoặc đong 180 ml nước cất) rồi trộn vào nhau và khuấy đều ta thu được 300 gam dung dịch \(Ba\left(OH\right)_2\) 10%

Tất cả viết đều sai nhé: )

Sửa:

A: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

B: \(S+O_2\underrightarrow{t^o}SO_2\)

C: \(2Fe+3Cl_2\rightarrow2FeCl_3\)

D: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Chọn D: CuO - đồng (II) oxit

FeO: sắt II oxit

`SO_2`: lưu huỳnh đioxit

a)

 \(-\dfrac{2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\\ \Rightarrow\dfrac{1}{3}\left(2x-5\right)=-\dfrac{2}{3}-\dfrac{3}{2}\\ \Rightarrow\dfrac{1}{3}\left(2x-5\right)=-\dfrac{13}{6}\\ \Rightarrow2x-5=-\dfrac{13}{6}:\dfrac{1}{3}=-\dfrac{13}{2}\\ \Rightarrow2x=-\dfrac{13}{2}+5\\ \Rightarrow2x=-\dfrac{3}{2}\\ \Rightarrow x=-\dfrac{3}{2}:2\\ \Rightarrow x=-\dfrac{3}{4}\)

b)

\(\dfrac{2}{5}x+\dfrac{1}{2}=-\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{5}x=-\dfrac{3}{4}-\dfrac{1}{2}\\ \Rightarrow\dfrac{2}{5}x=-\dfrac{5}{4}\\ \Rightarrow x=-\dfrac{5}{4}:\dfrac{2}{5}=-\dfrac{25}{8}\)