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

Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)=n_{ZnCl_2}=n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)

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

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

\(\Rightarrow m_{Cu}=0,1\cdot64=6,4\left(g\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2}=0,1.22,4=2,24\left(l\right)\)

c, Ta có: \(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

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

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

\(\Rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)

Bạn tham khảo nhé!

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

PTHH :

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

0,4      1,2           0,4         0,6 

\(a,V_{H_2}=0,6.22,4=13,44\left(l\right)\)

\(b,m_{HCl}=1,2.36,5=43,8\left(g\right)\)

\(c,m_{AlCl_3}=133,5.0,4=53,4\left(g\right)\)

\(m_{ddHCl}=\dfrac{43,8.100}{10}=438\left(g\right)\)

\(m_{ddAlCl_3}=10,8+438-\left(0,6.2\right)=447,6\left(g\right)\)

\(C\%=\dfrac{53,8}{447,6}.100\%\approx12,02\%\)

Bài này anh có hỗ trợ ở dưới rồi nha em!

nCuO=16/80=0,2(mol)

a) PTHH: CuO + H2SO4 -> CuSO4 + H2O

0,2___________0,2_____0,2(mol)

b) mCuSO4=160.0,2=32(g)

c) KL dung dịch H2SO4 tham gia p.ứ là đề bài cho 100(g)

d) mH2SO4=0,2.98=19,6(g)

=>C%ddH2SO4= (19,6/100).100=19,6%

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

a) PTHH: Fe + 2HCl --> FeCl₂ + H₂

_______0,2---->0,4----->0,2--->0,2

=> V_H2 = 0,2.22,4 = 4,48(l)

b) m_HCl = 0,4.36,5 = 14,6(g)

c) m_FeCl2 = 0,2.127 = 25,4 (g)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ 0,05.........0,1..........0,05..........0,05\left(mol\right)\\ a.C\%_{ddHCl}=\dfrac{0,1.36,5}{200}.100=1,825\%\\ b.m_{Zn}=0,05.65=3,25\left(g\right)\\ c.C\%_{ddZnCl_2}=\dfrac{136.0,05}{3,25+200-0,05.2}.100\approx3,347\%\)

n_Al = 5.4 / 27 = 0.2 (mol)

2Al + 6HCl => 2AlCl₃ + 3H₂

0.2......0.6............0.2.......0.3

a) V_H2 = 0.3 * 22.4 = 6.72 (l) 

b) m_AlCl3 = 0.2 * 133.5 = 26.7 (g) 

c) VddHCl = 0.6 / 1.5 = 0.4 (l) 

d) C_MAlCl3 = 0.2 / 0.4 = 0.5 (M) 

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

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{HCl}=\dfrac{0,6}{1,5}=0,4\left(l\right)=400\left(ml\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\end{matrix}\right.\)

\(pthh:Mg+2HCl--->MgCl_2+H_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\n_{HCl}=\dfrac{\dfrac{109,5.10\%}{100\%}}{36,5}=0,3\left(mol\right)\end{matrix}\right.\)

a. Ta thấy: \(\dfrac{0,2}{1}>\dfrac{0,3}{2}\)

Vậy Mg dư

Theo pt: \(n_{H_2}=\dfrac{1}{2}.0,3=0,15\left(mol\right)\)

\(\Rightarrow m_{H_2}=0,15.2=0,3\left(g\right)\)

b. \(m_{dd_{sau.PỨ}}=4,8+109,5-0,3=114\left(g\right)\)

c. Theo pt: \(n_{MgCl_2}=0,15\left(mol\right)\)

\(\Rightarrow C_{\%_{MgCl_2}}=\dfrac{0,15.95}{114}.100\%=12,5\%\)

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

m HCl=\(\dfrac{109,5.10}{100}\)=10,95g

=>n HCl=\(\dfrac{10,95}{36,5}\)=0,3 mol

Mg + 2HCl -> MgCl₂ + H₂

0,15---0,3---------0,15----0,15 mol

=>HCl phản ứng hết, Mg dư

=> m_Mg = 0,15.24 = 3,6 (g)

=> m = 0,15.2 = 0,3 (g)

m= m_Mg + m_ddHCl- m = 3,6 + 109,5 - 0,3 = 112,8 (g)

c)

 m= 0,15.95 = 14,25 (g)

->C%_ddMgCl2=\(\dfrac{14,25}{112,8}\)_.100=12,63%

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

\(m_{HCl}=\dfrac{109,5\cdot10\%}{100\%}=10,95g\Rightarrow n_{HCl}=0,3mol\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

0,1         0,3          0           0

0,1         0,2          0,1        0,1

0            0,1          0,1        0,1

a)\(V_{H_2}=0,1\cdot22,4=2,24l\)

b)\(m_{MgCl_2}=0,1\cdot95=9,5g\)

c)\(m_{H_2}=0,1\cdot2=0,2g\)

\(m_{ddMgCl_2}=4,8+109,5-0,2=114,1g\)

\(C\%=\dfrac{9,5}{114,1}\cdot1005=8,33\%\)

\(a.n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{109,5.10\%}{36,5}=0,3\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ LTL:\dfrac{0,2}{1}>\dfrac{0,3}{2}\\ \Rightarrow Mgdư\\ n_{H_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow m_{H_2}=0,15.2=0,3\left(g\right)\\ b.m_{ddsaupu}=m_{Mg\left(pứ\right)}+m_{ddHCl}-m_{H_2}=0,15.24+109,5-0,3=112,8\left(g\right)\\ c.n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow C\%_{MgCl_2}=\dfrac{0,15.95}{112,8}.100=12,63\%\)

mHCl = 109,5 . 10% = 10,95 (g)

nHCl = 10,95/36,5 = 0,3 (mol)

nMg = 4,8/24 = 0,2 (mol)

PTHH: Mg + 2HCl -> MgCl2 + H2

LTL: 0,2 > 0,3/2 => Mg dư

nH2 = 0,3/2 = 0,15 (mol)

mH2 = 0,15 . 2 = 0,3 (g)

mdd = 4,8 + 109,5 - 0,3 = 114 (g)

nMg (p/ư) = nMgCl2 = nH2 = 0,15 (mol)

mMg (dư) = (0,2 - 0,15) . 24 = 1,2 (g)

mMgCl2 = 0,2 . 95 = 19 (g)

C%Mg = 1,2/114 = 1,05%

C%MgCl2 = 19/114 = 16,66%