Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

a. PTHH:

\(Mg+2HCl--->MgCl_2+H_2\left(1\right)\)

\(MgO+2HCl--->MgCl_2+H_2O\left(2\right)\)

b. Theo PT₍₁₎: \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,2.24=4,8\left(g\right)\)

\(\Rightarrow m_{MgO}=8-4,8=3,2\left(g\right)\)

c. Ta có: \(n_{hh}=0,2+\dfrac{3,2}{40}=0,28\left(mol\right)\)

Theo PT_(1,2): \(n_{HCl}=2.n_{hh}=2.0,28=0,56\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,56.36,5=20,44\left(g\right)\)

ai làm giúp mình với

\(a.Mg+2HCl->MgCl_2+H_2\\ MgO+2HCl->MgCl_2+H_2O\\ b.n_{H_2}=\dfrac{2,24}{22,4}=n_{Mg}=0,1mol\\ \%m_{Mg}=\dfrac{0,1.24}{6}=40\%;\%m_{MgO}=60\%\\ n_{MgO}=\dfrac{0,6.6}{40}=0,09\left(mol\right)\\ n_{MgCl_2}=0,1+0,09=0,19\left(mol\right)\\ n_{HCl}=0,19.2=0,38\left(mol\right)\\ V_{ddHCl}=\dfrac{0,38.36,5}{0,2.1,1}=63,0\left(mL\right)\\ C_{M\left(MgCl_2\right)}=\dfrac{0,19}{0,063}=3,0\left(M\right)\)

\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

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

\(0.1.......0.2......................0.1\)

Chất rắn X : Cu 

\(m_{Zn}=0.1\cdot65=6.5\left(g\right)\Rightarrow m_{Cu}=19.3-6.5=12.8\left(g\right)\)

\(n_{Cu}=\dfrac{12.8}{64}=0.2\left(mol\right)\)

\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)

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

\(0.2........0.1\)

\(m_{tăng}=m_{O_2}=0.1\cdot32=3.2\left(g\right)\)

a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

                \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

b) Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{Zn}\) \(\Rightarrow n_{ZnO}=\dfrac{20-0,1\cdot65}{81}=\dfrac{1}{6}\left(mol\right)\)

\(\Rightarrow n_{ZnCl_2}=n_{Zn}+n_{ZnO}=\dfrac{4}{15}\left(mol\right)\) 

Mặt khác: \(m_{H_2}=0,1\cdot2=0,2\left(g\right)\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=119,8\left(g\right)\) \(\Rightarrow C\%_{ZnCl_2}=\dfrac{\dfrac{4}{15}\cdot136}{119,8}\cdot100\%\approx30,27\%\)

c) Giả sử khí là SO₂

PTHH: \(Zn+H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}ZnSO_4+SO_2\uparrow+H_2O\)

Theo PTHH: \(n_{SO_2}=n_{Zn}=0,1\left(mol\right)\) \(\Rightarrow V_{SO_2}=0,1\cdot22,4=2,24\left(l\right)\)

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

b, Ta có: 24n_Mg + 56n_Fe = 10,4 (1)

Theo PT: \(n_{H_2}=n_{Mg}+n_{Fe}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(2\right)\)

Từ (1)  và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,2\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,2.24=4,8\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\end{matrix}\right.\)

c, Theo PT: \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\Rightarrow V_{ddHCl}=\dfrac{0,6}{0,5}=1,2\left(l\right)\)

C1:

nH2= 0,15(mol)

PTHH: Fe + 2 HCl-> FeCl2 + H2

0,15_____0,3______0,15___0,15(mol)

-> mFe= 0,15.56=8,4(g)

-> mCuO= 10-8,4=1,6(g)

b) -> nCuO= 0,02(mol)

PTHH: CuO +2 HCl -> CuCl2 + H2O

0,02_________0,04(mol)

c) nHCl (tổng)= 0,34(mol)

=> CMddHCl= 0,34/0,2=1,7(M)

C2:

a) Mg +2 HCl -> MgCl2 + H2

nH2= 0,2(mol) -> nMg= nMgCl2= nH2=0,2(mol); nHCl=0,4(mol)

mMg= 0,2.24=4,8(g) -> mCu= 5,2(g)

=> %mMg=(4,8/10).100=48%

=>%mCu= 100%-48%=52%

b) mMgCl2= 0,2.95=19(g)

mHCl=0,4.36,5=14,6(g) -> mddHCl= ?? Không cho C% sao tính ta