\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{Zn}=n_{H_2}=\dfrac{3,7185}{24,79}=0.,15(mol)\\ \Rightarrow m_{Zn}=0,15.65=9,75(g)\\ \Rightarrow \%_{Zn}=\dfrac{9,75}{10}.100\%=97,5\%\\ \Rightarrow \%_{Cu}=100\%-97,5\%=2,5\%\\ b,n_{HCl}=2n_{H_2}=0,3(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,3.36,5}{14\%}=78,21(g)\)

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

\(R+2HCl\rightarrow RCl_2+H_2\)

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

\(M_R=\dfrac{13.7}{0.1}=137\left(\dfrac{g}{mol}\right)\)

\(R:Ba\)

\(200\left(ml\right)=0.2\left(l\right)\)

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

câu a

dạ em làm xong câu B rồi mọi người khỏi cần trả lời nữa ạ

a. PTHH: R + H₂SO₄ ---> RSO₄ + H₂ (1)

Ta có: \(n_{H_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)

Theo PT₍₁₎: \(n_R=n_{H_2}=0,16\left(mol\right)\)

=> \(M_R=\dfrac{3,84}{0,16}=24\left(g\right)\)

Vậy R là magie (Mg)

b. PTHH:

Mg + HCl ---> MgCl₂ + H₂ (2)

Theo PT₍₂₎: \(n_{H_2}=n_{Mg}=0,16\left(mol\right)\)

=> \(V_{H_2}=0,16.22,4=3,584\left(lít\right)\)

(Do câu b đề ko rõ lắm nên mik làm như vậy, nếu sai bn bình luận nhé.)

Mg+2HCl->MgCl2+H2

0,2-----------------------0,2

RO+H2-to>R+H2O

0,2-------------0,2

n Mg=\(\dfrac{4,8}{24}\)=0,2 mol

=>VH2=0,2.22,4=4,48l

->0,2=\(\dfrac{12,8}{R}\)

=>R=64 g\mol

=>R là Cu(đồng)

Bài 8:

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

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

_____0,2______0,6_____0,2____0,3 (mol)

a, \(m_{Al}=0,2.27=5,4\left(g\right)\)

b, \(C_{M_{HCl}}=\dfrac{0,6}{0,3}=2\left(M\right)\)

c, \(C_{M_{AlCl_3}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Bài 9:

Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

a, \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)

\(\Rightarrow m_{MgO}=8,4-2,4=6\left(g\right)\)

b, \(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}+2n_{MgO}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{3,65\%}==500\left(g\right)\)

a) nH2=0,15(mol)

PTHH: R + 2 HCl -> RCl2 + H2

0,15______0,3____0,15___0,15(mol)

M(R)=mR/nR=3,6/0,15= 24(g/mol)

=> R(II) cần tìm là Magie (Mg=24)

b) PTHH: Mg + 2 HCl -> MgCl2 + H2

mHCl=0,3.36,5=10,95(g)

=>C%ddHCl= (10,95/150).100= 7,3%

c) mH2= 0,15.2=0,3(g)

mddMgCl2= mMg + mddHCl - mH2= 3,6+ 150 - 0,3= 153,3(g)

mMgCl2=0,15.95=14,25(g)

=> \(C\%ddMgCl2=\dfrac{14,25}{153,3}.100\approx9,295\%\)