\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{Cl_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\\ a,V_{Cl_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,m_{FeCl_3}=162,5.0,2=32,5\left(g\right)\)

a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PT: \(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

Theo PT: \(n_{Cl_2}=\dfrac{3}{2}n_{Fe}=0,3\left(mol\right)\)

\(\Rightarrow V_{Cl_2}=0,3.22,4=6,72\left(l\right)\)

b, \(n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)

a/ \(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:     0,3      0,6         0,3       0,3

\(m_{Mg}=0,3.24=7,2\left(g\right)\)

b/ \(m_{MgCl_2}=0,3.95=28,5\left(g\right)\)

c/ \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)

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

Ta có: \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,1\left(mol\right)\\n_{CuCl_2}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,1\cdot36,5}{7,3\%}=50\left(g\right)\\C\%_{CuCl_2}=\dfrac{0,05\cdot135}{4+50}\cdot100\%=12,5\%\end{matrix}\right.\)

a)

\(2Cu + O_2 \xrightarrow{t^o} 2CuO\)

b)

\(n_{CuO} = n_{Cu} = \dfrac{6,4}{64} = 0,1(mol)\\ \Rightarrow m_{CuO} = 0,1.80 = 8(gam)\)

c)

\(n_{O_2} = \dfrac{1}{2}n_{Cu} = 0,05(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ m_{KMnO_4} = 2n_{O_2} = 0,05.2 = 0,1.158 = 15,8(gam)\)

d)

\(V_{không\ khí} = 5V_{O_2} = 0,05.22,4.5 = 5,6(lít)\)

Bài 1: Ta có: \(n_{Mg}=\dfrac{24}{24}=1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

____1_____2_____________1 (mol)

a, Ta có: \(V_{H_2}=1.22,4=22,4\left(l\right)\)

b, Ta có: \(m_{HCl}=2.36,5=73\left(g\right)\)

Bài 2: Ta có: \(n_{CuO}=\dfrac{100}{80}=1,25\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

___1,25_______1,25__1,25 (mol)

a, Ta có: \(m_{Cu}=1,25.64=80\left(g\right)\)

b, \(m_{H_2O}=1,25.18=22,5\left(g\right)\)

Bạn tham khảo nhé!

a) Đặt: nZn=x(mol); nFe= y(mol) (x,y: nguyên, dương)

Zn + H2SO4 -> ZnSO4 + H2

x_______x_______x________x

Fe + H2SO4 -> FeSO4 + H2

y____y_________y___y(mol)

b) m(rắn)=mCu=3(g)

=> m(Zn, Fe)= 21,6 - 3= 18,6(g)

Ta có hpt:

\(\left\{{}\begin{matrix}65x+56y=18,6\\22,4x+22,4y=6,72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

=> Zn= 65.0,2=13(g)

=>%mZn= (13/21,6).100=60,185%

%mCu=(3/21,6).100=13,889%

=>%mFe=25,926%

c) nH2SO4=x+y=0,3(mol) =>mH2SO4=29,4(g)

=> mddH2SO4= (29,4.100)/25=117,6(g)

a) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH:
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (1)

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\) (2)

Theo PT (1): \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Al_2O_3}=15,6-5,4=10,2\left(g\right)\end{matrix}\right.\)

b) \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

Theo PT (1), (2): \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}+n_{Al_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{mu\text{ố}i}=m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\)

c) Theo PT (1), (2): \(n_{H_2SO_4}=n_{H_2}+3n_{Al_2O_3}=0,6\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(c\text{ần}.d\text{ùng}\right)}=0,6.98=58,8\left(g\right)\)

PTHH: \(Cu+Cl_2\underrightarrow{t^o}CuCl_2\)

Ta có: \(m_{CuCl_2\left(lýthuyết\right)}=\dfrac{2,7}{80\%}=3,375\left(g\right)\) \(\Rightarrow n_{CuCl_2}=\dfrac{3,375}{135}=0,025\left(mol\right)=n_{Cu}=n_{Cl_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Cu}=0,025\cdot64=1,6\left(g\right)\\V_{Cl_2}=0,025\cdot22,4=0,56\left(l\right)\end{matrix}\right.\)