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

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

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

2Al + 6HCl -> 2AlCl₃ + 3H₂ (1)

ZnO + 2HCl -> ZnCl₂ + H₂O (2)

a) n_H2= 13,44/22.4=0.6(mol) -> mH2=0,6.2=1,2(g)

Theo PTHH: n_Al = 2/3 n_H2 = 2/3 . 0,6= 0,4(mol) -> m_Al = 0,4 . 27=10,8(g)

-> m_ZnO = 27-10,8= 16,2(g)

b) n_ZnO = 16,2/81=0,2(mol)

Theo PTHH (2): nHCl = 2nZnO=2.0,2=0,4(mol)

Theo PTHH (1) : nHCl=2nH₂=2.0,6=1,2(mol)

-> \(\Sigma\)nHCl = 0,4+1,2=1,6(mol)

-> mHCl = 1,6.36,5= 58,4(g)

-> mdd_HCl = 58,4.100/29,2= 200(g)

c) Theo PTHH (1): nAlCl₃ = 2/3 nH₂ = 2/3 . 0,6=0,4(mol)                                         -> mAlCl3=0,4.133,5=53,4(g)

mdd sau phản ứng= m_A + mddHCl - mH2 =27+200-1,2 =225,8(g)

-> C% AlCl3 = 53,4.100%/225,8 = 20,88%

Theo PTHH (2) nZnCl2 =nZnO= 0,2(mol)-> mZnCl2=0,2.136=27,2(g)

-> C% ZnCl2= 27,2.100%/255,8=10,63%

Tiếp bài của creeper nhé:

c. Ta có: \(n_{ZnO}=\dfrac{4,86}{81}=0,06\left(mol\right)\)

Theo PT₍₁₎: \(n_{HCl}=2.n_{ZnO}=2.0,06=0,12\left(mol\right)\)

Theo PT₍₂₎: \(n_{HCl}=2.n_{Zn}=2.0,1=0,2\left(mol\right)\)

=> \(n_{HCl}=0,12+0,2=0,32\left(mol\right)\)

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

Ta có: \(C_{\%_{HCl}}=\dfrac{11,68}{m_{dd_{HCl}}}.100\%=12\%\)

=> \(m_{dd_{HCl}}=\dfrac{292}{3}\left(g\right)\)

Theo đề, ta có: 

\(D=\dfrac{\dfrac{292}{3}}{V_{dd_{HCl}}}=1,2\)(g/ml)

=> \(V_{dd_{HCl}}=81,1\left(ml\right)\)

a, PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

Ta có: \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,4.27}{14,8}.100\%\approx72,97\%\\\%m_{MgO}\approx27,03\%\end{matrix}\right.\)

b, Ta có: \(n_{MgO}=\dfrac{14,8-0,4.27}{40}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{H_2}+n_{MgO}=0,7\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,7.98=68,6\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{68,6}{10\%}=686\left(g\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\\n_{MgSO_4}=n_{MgO}=0,1\left(mol\right)\end{matrix}\right.\)

Ta có: m _{dd sau pư} = 14,8 + 686 - 0,6.2 = 699,6 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,2.342}{699,6}.100\%\approx9,78\%\\C\%_{MgSO_4}=\dfrac{0,1.120}{699,6}.100\%\approx1,72\%\end{matrix}\right.\)

Zn+2HCl->ZnCl2+H2

ZnO+2HCl->ZnCl2+H2O

nH2=0.2(mol)->nZn=0.2(mol).mZn=13(g)

mZnO=21.1-13=8.1(g)

nZnO=0.1(mol)

Tổng nHCl cần dùng:0.2*2+0.1*2=0.6(mol)

mHCl=21.9(g)

C%ddHCl=21.9:200*100=10.95%

n muối=0.2+.1=0.3(mol)

m muối=40.8(g)

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

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