cho 13(g) kẽm tác dụng với dung dịch axit clohiric theo sơ đồ phản ứng sau:

Zn + HCl --> ZnCL2 + h2

a/ lập phương trình hóa học

b/ tính khối lượng axit clohidic cần dùng

c/ tính thể tích khí hydro thoát ra

d1/ chất nào dư sau phản ứng

d2/ thu được bao nhiêu gam đồng kim loại

CHÚC BẠN HỌC TỐT! ^^

a, nAl = 5,4/27 = 0,2 ( mol )

PTHH: (1) 2Al + 6HCl ---> 2AlCl3 + 3H2

b, Theo PT (1) :

nHCl = 3nAl = 0,2.3 = 0,6 ( mol )

=> mHCl = 0,6.36,5=21,9 ( g )

=> C%HCl = (21,9.100)/400 = 5,475%

c, Theo PT (1) :

nH2 = 3/2.nAl = 3/2.0,2 = 0,3 ( mol )

=> mH2 = 0,3.2 = 0,6 ( g )

Áp dụng định luật bảo toàn khối lượng

=> mAlCl3 + mH2 = mAl + mHCl

=> mAlCl3 = mAl + mHCl - mH2= 304,8(g)

Theo PT (1) :

nAlCl3 = nAl = 0,2 ( mol )

=> mAlCl3 = 0,2.133 =26,7 ( g )

=>C% các chất sau pứ= (26,7.100)/304,8 = 8,8%

\(n_{Na_2O}=\dfrac{12,4}{62}=0,2mol\\ a,Na_2O+H_2O\rightarrow2NaOH\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{HCl}=n_{NaCl}=0,2.2=0,4mol\\ b,C_{M_{HCl}}=\dfrac{0,4}{0,2}=2M\\ c,m_{ddNaCl}=200.1,05+12,4=222,4g\\ C_{\%NaCl}=\dfrac{0,4.58,5}{222,4}\cdot100=10,52\%\)

a) PT phân tử: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)

    PT ion: \(CaCO_3+2H^+\rightarrow Ca^{2+}+H_2O+CO_2\uparrow\)

b) Ta có: \(\left\{{}\begin{matrix}n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\\n_{HCl}=\dfrac{43,8\cdot20\%}{36,5}=0,24\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,24}{2}\) \(\Rightarrow\) HCl dư, tính theo CaCO₃

\(\Rightarrow\left\{{}\begin{matrix}n_{CaCl_2}=0,1\left(mol\right)=n_{CO_2}\\n_{HCl\left(dư\right)}=0,04\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCl_2}=0,1\cdot111=11,1\left(g\right)\\m_{CO_2}=0,1\cdot44=4,4\left(g\right)\\m_{HCl\left(dư\right)}=0,04\cdot36,5=1,46\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{CaCO_3}+m_{ddHCl}-m_{CO_2}=49,4\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{11,1}{49,4}\cdot100\%\approx22,47\%\\C\%_{HCl\left(dư\right)}=\dfrac{1,46}{49,4}\cdot100\%\approx2,96\%\end{matrix}\right.\)

Bổ sung: \(D_{HCl}=1,18\left(g/ml\right)\)

a) PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

b) Ta có: \(\left\{{}\begin{matrix}n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\\n_{HCl}=\dfrac{100\cdot1,18\cdot20\%}{36,5}=\dfrac{236}{365}\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{\dfrac{236}{365}}{2}\) \(\Rightarrow\) HCl còn dư, MgO p/ứ hết

\(\Rightarrow n_{MgCl_2}=0,05\left(mol\right)\) \(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)

\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ a,PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=0,05\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ V_{H_2\left(đkc\right)}=0,05.24,79=1,2395\left(l\right)\\ b,n_{HCl}=0,1.3=0,3\left(mol\right)\\ Vì:\dfrac{0,05}{1}< \dfrac{0,3}{2}\Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,3-0,05.2=0,2\left(mol\right)\\ n_{FeCl_2}=n_{Fe}=0,05\left(mol\right)\\ V_{ddsau}=V_{ddHCl}=0,1\left(l\right)\\ b,C_{MddFeCl_2}=\dfrac{0,05}{0,1}=0,5\left(M\right);C_{MddHCl\left(dư\right)}=\dfrac{0,2}{0,1}=2\left(M\right)\)

nAl= 0,5(mol)

a) PTHH: 2 Al + 6 HCl -> 2 AlCl3 + 3 H2

nHCl= 6/2 . 0,5= 1,5(mol)

=>mHCl= 1,5.36,5=54,75(mol)

=> mddHCl= (54,75.100)/18,25=300(g)

b) nH2= 3/2. 0,5=0,75(mol)

=>V(H2,đktc)=0,75.22,4=16,8(l)

c) nAlCl3= nAl= 0,5(mol) -> mAlCl3=0,5. 133,5=66,75(g)

mddAlCl3=mAl+ mddHCl - mH2= 13,5 + 300-0,75.2=312(g)

=> \(C\%ddAlCl3=\dfrac{66,75}{312}.100\approx21,394\%\)

a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)

c, Ta có: m _{dd sau pư} = 8,1 + 200 - 0,45.2 = 207,2 (g)

Theo PT: \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{0,3.133,5}{207,2}.100\%\approx19,33\%\)

Câu 1 : 

\(n_{Mg}=\dfrac{8.4}{24}=0.35\left(mol\right)\)

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

\(0.35.......0.7.........0.35..........0.35\)

\(C\%_{HCl}=\dfrac{0.7\cdot36.5}{146}\cdot100\%=17.5\%\)

\(m_{\text{dung dịch sau phản ứng}}=8.4+146-0.35\cdot2=153.7\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{0.35\cdot95}{153.7}\cdot100\%=21.6\%\)

Câu 2 :

\(n_{CaCO_3}=\dfrac{10}{100}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{114.1\cdot8\%}{36.5}=0.25\left(mol\right)\)

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

\(1................2\)

\(0.1.............0.25\)

\(LTL:\dfrac{0.1}{1}< \dfrac{0.25}{2}\Rightarrow HCldư\)

\(m_{\text{dung dịch sau phản ứng}}=10+114.1-0.1\cdot44=119.7\left(g\right)\)

\(C\%_{HCl\left(dư\right)}=\dfrac{\left(0.25-0.2\right)\cdot36.5}{119.7}\cdot100\%=1.52\%\)

\(C\%_{CaCl_2}=\dfrac{0.2\cdot111}{119.7}\cdot100\%=18.54\%\)

Câu 3 : 

\(n_{H_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0.3.......0.6......................0.3\)

\(m_{Fe}=0.3\cdot56=16.8\left(g\right)\)

\(m_{dd_{HCl}}=150\cdot1.2=180\left(g\right)\)

\(C\%_{HCl}=\dfrac{0.6\cdot36.5}{180}\cdot100\%=12.16\%\)

\(C_{M_{HCl}}=\dfrac{0.6}{0.15}=4\left(M\right)\)