\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ ZnO+2HCl\to ZnCl_2+H_2O\\ b,n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ \Rightarrow n_{Zn}=0,15(mol)\Rightarrow m_{Zn}=0,15.65=9,75(g)\\ \Rightarrow \%_{Zn}=\dfrac{9,75}{25,95}.100\%=37,57\%\\ \Rightarrow \%_{ZnO}=(100-37,57)\%=62,43\%\\ c,n_{ZnO}=\dfrac{25,95-9,75}{81}=0,2(mol)\\ \Rightarrow n_{HCl}=2.0,15+2.0,2=0,7(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,7.36,5}{12\%}=212,92(g)\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1=n_{Zn}\\ n_{ZnO}=\dfrac{14,6-6,5}{81}=0,1mol\\ C\%=\dfrac{0,2\cdot136}{175,6+14,6-0,2}=14,32\%\)

n_H2 = \(\dfrac{2,24}{22,4}=0,1\) mol

Pt: ZnO + HCl --> ZnCl₂ + H₂O

.....Zn + 2HCl --> ZnCl₂ + H₂

0,1 mol<--------------------0,1 mol

m_Zn = 0,1 . 65 = 6,5 (g)

m_ZnO = m_hh - m_Zn = 26 - 6,5 = 19,5 (g)

=> n_ZnO = \(\dfrac{19,5}{81}\) mol

=> n_Zn = n_ZnO = \(\dfrac{19,5}{81}\) mol

=> m_Zn = \(\dfrac{19,5}{81}.65=15,648\left(g\right)\)

\(\sum m_{Zn}=6,5+15,648=22,148\left(g\right)\)

% m_Zn = \(\dfrac{22,148}{26}.100\%=85,2\%\)

\(n_{HCl} = \dfrac{448.1,12.3,65\%}{36,5} = 0,50176(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ ZnO + 2HCl \to ZnCl_2 + H_2O\\ n_{Zn} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)\\ n_{ZnO} = \dfrac{n_{HCl} - 2n_{Zn}}{2} = \dfrac{0,50176-0,1.2}{2} = 0,15088(mol)\\ \%m_{Zn} = \dfrac{0,1.65}{0,1.65 + 0,15088.81}.100\% = 34,72\%\\ \%m_{ZnO} = 65,28\%\)

Giải thích: 

Zn + 2HCl -> ZnCl₂ + H₂

ZnO + 2HCl -> ZnCl₂ + H₂O

n_H2 = n_Zn = 0,4 mol

n_HCl = 1,2 mol = 2n_Zn + 2n_ZnO => n_ZnO = 0,2 mol

=> %m_ZnO = 38,4%

Đáp án B

nH2 = 2.24/22.4 = 0.1 (mol) 

Zn + 2HCl => ZnCl2 + H2 

ZnO + 2HCl => ZnCl2 + H2O 

nZn = nH2 = 0.1 (mol) 

=> mZn = 6.5 g

mZnO = 14.6 - 6.5 = 8.1 (g) 

nZnO = 0.1 (mol)  

%Zn = 6.5/14.6 * 100% = 44.52%

%ZnO = 55.48%

nHCl = 0.1*2 + 0.1*2= 0.4 (mol) 

Vdd HCl = 0.4 / 0.5 = 0.8 l 

\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

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

\(0.2.........0.4.........0.2......0.2\)

\(m_{Zn}=0.2\cdot65=13\left(g\right)\Rightarrow m_{ZnO}=14.6-13=1.6\left(g\right)\)

\(\%Zn=\dfrac{13}{14.6}\cdot100\%=89.04\%\)

\(\%ZnO=100\%-89.04\%=10.96\%\)

\(n_{ZnO}=\dfrac{1.6}{81}\approx0.02\left(mol\right)\)

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

\(0.02........0.04........0.02........0.02\)

\(n_{HCl}=0.4+0.04=0.44\left(mol\right)\)

\(C_{M_{HCl}}=\dfrac{0.44}{0.8}=0.55\left(M\right)\)

Eeeee ngồi tính sang chấn thật nó ra số xấu lần mò hơn 20p chưa biết tính sai chỗ nào

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

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