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

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%

a) $2Al + 6HCl \to 2AlCl_3 + 3H_2$

n Al = 5,4/27 = 0,2(mol)

Theo PTHH : n H2 = 3/2 n Al = 0,3(mol)

=> V H2 = 0,3.22,4 = 6,72(lít)

b) n HCl = 3n Al = 0,6(mol)

=> mdd HCl = 0,6.36,5/20% = 109,5 gam

c)Sau phản ứng,

mdd = m Al + mdd HCl - m H2 = 5,4 + 109,5 - 0,3.2 = 114,3(gam)

=> C% AlCl3 = 0,2.133,5/114,3  .100% = 23,36%

Thấy @Tuyết Nhi Melody nói mình ít hoạt động quá! Giaỉ giúp bạn bài này nha!

Bài làm:

Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ m_{HCl}=\dfrac{109,5.10}{100}=10,95\left(g\right)\\ =>n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHHL Mg + 2HCl -> MgCl₂ + H₂

Ta có: \(\dfrac{0,2}{1}>\dfrac{0,3}{2}\)

=> Mg dư, HCl hết nên tính theo n_HCl.

a) Ta có: \(n_{H_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ =>V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)

b và c) - Chất có trong dd sau khi f.ứ kết thúc là MgCl₂.

Ta có: \(n_{Mg\left(f.ứ\right)}=n_{MgCl_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ =>m_{Mg\left(f.ứ\right)}=0,15.24=3,6\left(g\right)\\ m_{MgCl_2}=95.0,15=14,25\left(g\right)\\ m_{H_2}=0,15.2=0,3\left(g\right)\\ =>m_{dd-sau-f.ứ}=3,6+109,5-0,3=112,8\left(g\right)\\ =>C\%_{dd-sau-f.ứ}=\dfrac{14,25}{112,8}.100\approx12,633\%\)

a) PTHH: Mg + 2HCl ----> MgCl₂ + H₂

n_Mg = \(\dfrac{4,8}{24}=0,2\left(mol\right)\)

m_HCl = \(\dfrac{109,5.10\%}{100\%}=10,95\left(g\right)\)

=> n_HCl = \(\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

Ta có tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,3}{2}\)=> HCl phản ứng hết, Mg dư

Theo PTHH: n_\(Mg\) = \(\dfrac{1}{2}n_{HCl}\) = 0,15 (mol)

=> m_Mg = 0,15.24 = 3,6 (g)

Theo PTHH: n_{\(H_2\)} = n_Mg = 0,15 (mol)

=> m_{\(H_2\)} = 0,15.2 = 0,3 (g)

=> m_\(ddspu\) = m_Mg + m_{\(dd_{HCl}\)} - m_{\(H_2\)} = 3,6 + 109,5 - 0,3 = 112,8 (g)

c) Theo PTHH: n_{\(MgCl_2\)} = n_Mg = 0,15 (mol)

=> m_{\(MgCl_2\)}= 0,15.95 = 14,25 (g)

=> C%_{\(ddMgCl_2\)}= \(\dfrac{14,25}{112,8}.100\%\approx12,63\%\)

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!

a)\(m_{ddHCl}=\dfrac{4,8}{10\%}.100\%=48\left(g\right)\)

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

\(PTHH:Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\)

\(n_{MgCl_2}=n_{Mg}=n_{H_2}=0,2\)

\(m_{MgCl_2}=0,2.95=19\left(g\right)\)

\(V_{H_2}=0,2.22,4=4.48\left(l\right)\)

c)\(m_{H_2}=0,2.2=0,4\left(g\right)\)

\(m_{ddMgCl_2}=4,8+48-0,4=52,4\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{19}{52,4}.100\%=36\%\)

nH₂ = \(\dfrac{8,96}{22,4}=0,4mol\)

Do chỉ có Mg tác dụng với HCl

Bảo toàn e

2nMg = 2nH₂ = 0,8 mol 

=> nMg = 0,4 mol

=> mMg = 9,6g

=> %mMg= \(\dfrac{9,6}{28,8}.100\%=33,33\%\)

%mCu = 100 - 33,33% = 66,67%

\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

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

\(n_{Mg}=0.4\left(mol\right)\)

\(\%Mg=\dfrac{0.4\cdot24}{28.8}\cdot100\%=33.33\%\)

\(\%Cu=100-33.33=66.67\%\)

\($a/$\\ Zn+2HCl \to ZnCl_2+H_2\\ b/\\ n_{Zn}=0,1(mol)\\ n_{HCl}=0,1.2=0,2(mol)\\ m_{ddHCl}=\frac{0,2.36,5.100}{7,3}=100(g)\\ C\%_{ZnCl_2}=\frac{0,1.136}{100+6,5-0,1.2}.100\%=12,8\%\)

\(a/\\ Zn+2HCl \to ZnCl_2+H_2\\ b/\\ n_{Zn}=\frac{6,5}{65}=0,1(mol)\\ n_{HCl}=0,2(mol)\\ m_{ddsaupu}=\frac{0,1.36,5.100}{7,3}+6,5-0,1.2=56,3(g)\\ C\%_{ZnCl_2}=\frac{0,1.136}{56,3}.100\%=24,16\%\)

a,\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,1    0,05       0,1

b, \(m_{HCl}=0,05.36,5=2,92\left(g\right)\)

  \(m_{ddHCl}=\dfrac{2,92}{7,3\%}=40\left(g\right)\)

  \(m_{ddsaupứ}=6,5+40-0,1.2=46,3\left(g\right)\)

\(\Rightarrow C\%_{ddZnCl_2}=\dfrac{0,1.136.100\%}{46,3}=29,4\%\)

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

PTHH: Mg + 2HCl --> MgCl₂ + H₂

_____0,1<-0,2------<0,1<---0,1

=> m_MgCl2 = 0,1.95 = 9,5 (g)

b) \(C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\)