a) nH2=0,2(mol)

PTHH: Mg + 2 HCl -> MgCl2 + H2

0,2______0,4_______0,2_____0,2(mol)

PTHH: MgO +2 HCl -> MgCl2 + H2O

b) mMgO= 8,8 - 0,2.24=4(g)

%mMgO= (4/8,8).100= 45,455%

=>%Mg=54,545%

c) nHCl(tổng)= 2. nMg + 2. nMgO= 2. 0,2+ 0,1.2=0,6(mol)

=> mHCl= 0,6.36,5=21,9(g)

=>mddHCl=(21,9.100)/7,3=300(g)

d) mddMgCl2= mddHCl + m(hỗn hợp ban đầu) - mH2

<=>mddHCl= 300+ 8,8- 0,2.2= 308,4(g)

nMgCl2=0,3(mol) => mMgCl2= 0,3.95=28,5(g)

=>C%ddMgCl2= (28,5/308,4).100=9,241%

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

a. PTHH: 

Mg + 2HCl ---> MgCl₂ + H₂ (1)

MgO + 2HCl ---> MgCl₂ + H₂O (2)

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

=> \(m_{Mg}=0,1.24=2,4\left(g\right)\)

=> \(m_{MgO}=4,4-2,4=2\left(g\right)\)

b. Ta có: \(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)

=> \(n_{hh}=0,05+0,1=0,15\left(mol\right)\)

Theo PT_(1,2): \(n_{HCl}=2.n_{hh}=2.0,15=0,3\left(mol\right)\)

=> \(V_{dd_{HCl}}=\dfrac{0,3}{0,4}=0,75\left(lít\right)\)

_{Pư: Mg+2HCl -> MgCl2 +H2}

_{MgO+2HCl-> MgCl2 +H2O}

_{MgCl2+2NaOH-> Mg(OH)2}↓_+2NaCl

_{Mg(OH)2-> MgO+H2O}

nMgO=14/40=0.35 mol

-> nMg(OH)2=nMgCl2=nMgO=0.35 mol

-> nMgO=nMg=nMgCl2=0.35 mol

gọi x,y lần lượt là số mol của Mg,MgO

->mhh=24x+40y=2 (1)

nMgCl2=x+y=0.35 (2)

giải hệ (1)(2) có x=0.103125

y=0.246875

- khối lượng của bạn nhập sai vì klg hỗn hợp quá nhỏ

a)\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)

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

0,15    0,3                        0,15

\(m_{Zn}=0,15\cdot65=9,75\left(g\right)\)

\(\%m_{Zn}=\dfrac{9,75}{17,85}\cdot100\%=54,62\%\)

\(\%m_{ZnO}=100\%-54,62\%=45,38\%\)

b)\(m_{ZnO}=17,85-9,75=8,1\left(g\right)\Rightarrow n_{ZnO}=\dfrac{8,1}{81}=0,1mol\)

  \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

  0,1         0,2

  \(\Rightarrow\Sigma n_{HCl}=0,3+0,2=0,5mol\)

  \(\Rightarrow V=\dfrac{0,5}{1}=0,5l=500ml\)

Tiếp bài của creeper nhé:

c. Ta có: \(n_{ZnO}=\dfrac{4,86}{81}=0,06\left(mol\right)\)

Theo PT₍₁₎: \(n_{HCl}=2.n_{ZnO}=2.0,06=0,12\left(mol\right)\)

Theo PT₍₂₎: \(n_{HCl}=2.n_{Zn}=2.0,1=0,2\left(mol\right)\)

=> \(n_{HCl}=0,12+0,2=0,32\left(mol\right)\)

=> \(m_{HCl}=0,32.36,5=11,68\left(g\right)\)

Ta có: \(C_{\%_{HCl}}=\dfrac{11,68}{m_{dd_{HCl}}}.100\%=12\%\)

=> \(m_{dd_{HCl}}=\dfrac{292}{3}\left(g\right)\)

Theo đề, ta có: 

\(D=\dfrac{\dfrac{292}{3}}{V_{dd_{HCl}}}=1,2\)(g/ml)

=> \(V_{dd_{HCl}}=81,1\left(ml\right)\)

\(a.PTHH:\)

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

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

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

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

\(\Rightarrow m_{MgO}=12-0,2.24=7,2\left(g\right)\)

\(\Rightarrow\%_{MgO}=\dfrac{7,2}{12}.100\%=60\%\)

c. Ta có: \(n_{hh}=0,2+\dfrac{7,2}{40}=0,38\left(mol\right)\)

Theo PT_(1,2): \(n_{HCl}=2.n_{hh}=2.0,38=0,76\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,76.36,5=27,74\left(g\right)\)

\(\Rightarrow m_{dd_{HCl}}=138,7\left(g\right)\)

\(\Rightarrow V_{dd_{HCl}}=126\left(ml\right)\)

Sửa đề : 200 ml dung dịch HCl

\(n_{CO_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(n_{HCl}=0.2\left(mol\right)\)

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)

\(0.05..............0.1.......0.05.........0.05\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(0.05.....0.2-0.1......0.05\)

\(m_{hh}=0.05\cdot84+0.05\cdot40=6.2\left(g\right)\)

\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)

\(n_{Mg\left(OH\right)_2}=n_{MgCl_2}=0.1\left(mol\right)\)

\(m_{Mg\left(OH\right)_2}=0.1\cdot58=5.8\left(g\right)\)