\(Mg+2HCl \to MgCl_2+H_2\\ n_{H_2}=0,15(mol)\\ \to n_{Mg}=n_{H_2}=0,15(mol)\\ \%m_{Mg}=\frac{0,15.24}{10}.100\%=36\%\\ \%m_{Cu}=100\%-36\%=64\%\)

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

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

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

a) nH2 = 6,72 : 22,4 = 0,3 (MOL) 

PTHH:

Zn + 2HCl → ZnCl2 + H2

x                         x          x (mol)

Mg + 2HCl → MgCl2 + H2

y                          y            y (mol)
ta có 
65x + 24y = 11,3 
x+y=0,3 
=> x = 0,1 (mol) 
=> y = 0,2 (mol)
=> mMg = 0,2 . 24 = 4,8 (G)
=> %mMg = \(\dfrac{4,8}{11,3}\) . 100% = 42,47 % 
=> %mZn = 100% - 42,47% = 57,53 %

Gọi x,y lần lượt là số mol của Zn, Mg

n_H2 =\(\dfrac{6,72}{22,4}\)=0,3 mol

Pt: Zn + 2HCl --> ZnCl₂ + H₂

.....x......................................x

....Mg + 2HCl --> MgCl₂ + H₂

.....y.......................................y

Ta có hệ pt: {65x+24y=11,3

                    x+y=0,3

  ⇔{x=0,1y=0,2

%m_Zn = 0,2×24\

a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)  (1)

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

b) Ta có: \(\Sigma n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Gọi số mol của Mg là \(a\) \(\Rightarrow n_{H_2\left(1\right)}=a\)

Gọi số mol của Zn là b \(\Rightarrow n_{H_2\left(2\right)}=b\)

Ta lập được hệ phương trình:

\(\left\{{}\begin{matrix}a+b=0,4\\24a+65b=17,8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,2\cdot24}{17,8}\cdot100\%\approx26,97\%\\\%m_{Zn}\approx73,03\%\end{matrix}\right.\)

a) Sửa đề: dd H₂SO₄ 9,8%

Ta có: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\) \(\Rightarrow m_{H_2}=0,35\cdot2=0,7\left(g\right)\)

Bảo toàn nguyên tố: \(n_{H_2SO_4}=n_{H_2}=0,35\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,35\cdot98}{9,8\%}=350\left(g\right)\)

\(\Rightarrow m_{dd}=m_{KL}+m_{H_2SO_4}-m_{H_2}=361,6\left(g\right)\)

b) Tương tự câu a

Bài 1:

\(n_{HCl}=2.0,16=0,32\left(mol\right);n_{H_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

PTHH: Fe + 2HCl → FeCl₂ + H₂

\(m_{H_2}=0,16.2=0,32\left(g\right)\)

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

Theo ĐLBTKL ta có: \(m_{MgCl_2+FeCl_2}=1,4+11,68-0,32=12,76\left(g\right)\)

Bài 12:

Theo ĐLBTKL, ta có:

\(m_{hhkl}+m_{O_2}=m_{hh.oxit}\\ \Leftrightarrow11,9+m_{O_2}=18,3\\ \Leftrightarrow m_{O_2}=18,3-11,9=6,4\left(g\right)\\ n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

\(m_O=\dfrac{15.12,8}{100}=1,92\left(g\right)\)

=> \(n_O=\dfrac{1,92}{16}=0,12\left(mol\right)\)

=> \(n_{H_2O}=0,12\left(mol\right)\)

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

Bảo toàn H: n_HCl = 0,12.2 + 0,15.2 = 0,54 (mol)

=> n_Cl = 0,54 (mol)

m_muối = m_{hh rắn} - m_O + m_Cl

= 15 - 1,92 + 0,54.35,5 = 32,25 (g)

Chọn C

BTNT H: nH₂SO₄ = nH₂ = 0,15 mol => m (dd H₂SO₄ 10%)  = 0,15.98.(100/10) = 147 gam

KL + dd H₂SO₄ → muối + H₂

BTKL: m muối = mKL + m (dd H₂SO₄ 10%)– mH₂ = 5,2 + 147 – 0,15.2 = 151,9 gam