Đặt: \(n_{Zn}=a\left(mol\right);n_{ZnO}=b\left(mol\right)\left(a,b>0\right)\)

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

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ ZnO+2HCl\rightarrow ZnCl_2+H_2O\\ \Rightarrow\left\{{}\begin{matrix}65a+81b=14,6\\a=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ b.m_{Zn}=0,1.65=6,5\left(g\right)\\ m_{ZnO}=0,1.81=8,1\left(g\right)\\ d.m_{ddHCl}=\dfrac{\left(0,1+0,1\right).2.36,5.100}{7,3}=200\left(g\right)\)

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

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

\(n_{Zn}=n_{H_2}=\dfrac{0.448}{22.4}=0.02\left(mol\right)\)

\(\Rightarrow m_{Zn}=0.02\cdot65=1.3\left(g\right)\)

\(m_{ZnO}=4.54-1.3=3.24\left(g\right)\)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{10}.100\%=65\%\\\%m_{ZnO}=35\%\end{matrix}\right.\)

c, \(n_{HCl}=0,1.0,5=0,05\left(mol\right)\)

\(n_{NaOH}=0,03.1=0,03\left(mol\right)\)

PT: \(HCl+NaOH\rightarrow NaCl+H_2O\)

Xét tỉ lệ: \(\dfrac{0,05}{1}>\dfrac{0,03}{1}\), ta được HCl dư.

→ Quỳ tím chuyển đỏ do acid dư.

nH2=0,1(mol)

PTHH: Mg + 2 HCl -> MgCl2 + H2

0,1__________0,2___________0,1(mol)

MgO + 2 HCl -> MgCl2 + H2O

0,05____0,1___0,05(mol)

mMg=0,1. 24= 2,4(g) -> mMgO=4,4-2,4= 2(g) -> nMgO=0,05((mol)

b) %mMg= (2,4/4,4).100=54,545%

=> %mMgO=45,455%

c) nHCl=0,3(mol) -> mHCl=0,3.36,5=10,95(g)

=> mddHCl=(10,95.100)/7,3=150(g)

a) Gọi n Zn = a(mol) ; n ZnO =  b(mol)

=> 65a + 81b = 14,6(1)

$Zn + 2HCl \to ZnCl_2 + H_2$
$ZnO + 2HCl \to ZnCl_2 + H_2O$

n ZnCl2 = a + b = 27,2/136 = 0,2(2)

Từ (1)(2) suy ra : a = b = 0,1

%m Zn = 0,1.65/14,6  .100% = 44,52%
%m ZnO = 100% -44,52% = 55,45%

b)

n HCl = 2n Zn + 2n ZnO = 0,4(mol)

m dd HCl = 0,4.36,5/7,3% = 200(gam)

\(n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Pt : \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2|\)

         1           1              1           1

         0,1       0,1                         0,1

         \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O|\)

           1             1              1                1

          0,1          0,1

a) \(n_{Zn}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(m_{Zn}=0,1.65=6,5\left(g\right)\)

\(m_{ZnO}=14,6-6,5=8,1\left(g\right)\)

b) Có : \(m_{MgO}=8,1\left(g\right)\)

\(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)

\(n_{H2SO4\left(tổng\right)}=0,1+0,1=0,2\left(mol\right)\)

\(V_{ddH2SO4}=\dfrac{0,2}{2}=0,1\left(l\right)\)

 Chúc bạn học tốt

C32: 

a, \(n_C=\dfrac{2,4}{12}=0,2\left(mol\right)\)

PT: \(C+O_2\underrightarrow{t^o}CO_2\)

Theo PT: \(n_{CO_2}=n_C=0,2\left(mol\right)\Rightarrow V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

b, \(n_{NaOH}=0,3.1=0,3\left(mol\right)\)

\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=1,5\) → Pư tạo NaHCO₃ và Na₂CO₃

PT: \(CO_2+NaOH\rightarrow NaHCO_3\)

\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=n_{NaHCO_3}+n_{Na_2CO_3}=0,2\\n_{NaOH}=n_{NaHCO_3}+2n_{Na_2CO_3}=0,3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{NaHCO_3}=0,1\left(mol\right)\\n_{Na_2CO_3}=0,1\left(mol\right)\end{matrix}\right.\)

⇒ m_NaHCO3 = 0,1.84 = 8,4 (g)

m_Na2CO3 = 0,1.106 = 10,6 (g)

c, \(C_{M_{NaHCO_3}}=C_{M_{Na_2CO_3}}=\dfrac{0,1}{0,3}=\dfrac{1}{3}\left(M\right)\)

Lần sau bạn đăng tách câu hỏi ra nhé.

C31:

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{14,6}.100\%\approx44,52\%\\\%m_{ZnO}\approx55,48\%\end{matrix}\right.\)

c, \(n_{ZnO}=\dfrac{14,6-0,1.65}{81}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Zn}+2n_{ZnO}=0,4\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,4.36,5}{10\%}=146\left(g\right)\)

Lượng khí hidro thoát ra là 0,15 mol.

a) Al₂O₃ + 2NaOH \(\rightarrow\) 2NaAlO₂ + H₂O

2Al + 2NaOH + 2H₂O \(\rightarrow\) 2NaAlO₂ + 3H₂.

b) Lượng Al phản ứng bằng 2/3 số mol khí hidro thoát ra và bằng 0,1 mol.

\(\Rightarrow\) %m_nhôm=\(\dfrac{27.0,1}{12,9}\).100%\(\approx\)20,93%, %m_{nhôm oxit}\(\approx\)100%-20,93%\(\approx\)79,07%.