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a, Ta có: 24nMg + 56nFe = 9,2 (g) (1)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
BT e, có: 2nMg + 2nFe = 2nH2 = 0,5 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,15\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,15.24}{9,2}.100\%\approx39,13\%\\\%m_{Fe}\approx60,87\%\end{matrix}\right.\)
b, BTNT H, có: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)
a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: R + 2HCl --> RCl2 + H2
____0,15<-----------------0,15
=> \(M_R=\dfrac{3,6}{0,15}=24\left(Mg\right)\)
b)
PTHH: Mg + 2HCl --> MgCl2 + H2
__________0,3<-----0,15<---0,15
=> \(V=\dfrac{0,3}{2}=0,15\left(l\right)=150\left(ml\right)\)
\(C_{M\left(MgCl_2\right)}=\dfrac{0,15}{0,15}=1M\)
\(a.n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:R+2HCl\rightarrow RCl_2+H_2\\ \Rightarrow n_R=n_{H_2}=0,15\left(mol\right)\\ \Rightarrow M_R=\dfrac{m_R}{n_R}=\dfrac{3,6}{0,15}=24\\ \)
\(\Rightarrow R\) là \(Magie\left(Mg\right)\)
\(b.n_{HCl}=2.n_{H_2}=2.0,15=0,3\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{n_{HCl}}{C_M}=\dfrac{0,3}{2}=0,15\left(l\right)=150ml\)
\(n_{MgCl_2}=n_{Mg}=0,15\left(mol\right)\\ \Rightarrow C_{M_{ddMgCl_2}}=\dfrac{n_{MgCl_2}}{V}=\dfrac{0,15}{0,15}=1\left(M\right)\)
Câu 1:
Gọi : nMg=a(mol); nMgO=b(mol) (a,b>0)
a) PTHH: Mg + 2 HCl -> MgCl2 + H2
a________2a_______a______a(mol)
MgO +2 HCl -> MgCl2 + H2O
b_____2b_______b___b(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}24a+40b=8,8\\22,4a=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
=> mMg=0,2.24=4,8(g)
=>%mMg= (4,8/8,8).100=54,545%
=> %mMgO= 45,455%
b) m(muối)=mMg2+ + mCl- = 0,3. 24 + 0,6.35,5=28,5(g)
c) V=VddHCl=(2a+2b)/2=0,3(l)=300(ml)
Câu 2:
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{HCl}=0,4\cdot2=0,8\left(mol\right)\end{matrix}\right.\)
PTHH: \(Ca+2HCl\rightarrow CaCl_2+H_2\uparrow\)
0,2____0,4_____0,2____0,2 (mol)
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)
0,2____0,4______0,2____0,2 (mol)
Ta có: \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,2\cdot40}{0,2\cdot40+0,2\cdot56}\cdot100\%\approx41,67\%\\\%m_{CaO}=58,33\%\\m_{CaCl_2}=\left(0,2+0,2\right)\cdot111=44,4\left(g\right)\end{matrix}\right.\)
a,\(n_{hhA}=\dfrac{56}{0,112}=500\left(mol\right)\)
b,Ta có: \(\dfrac{n_{N_2}}{1}=\dfrac{n_{H_2}}{4}=\dfrac{n_{N_2}+n_{H_2}}{1+4}=\dfrac{500}{5}=100\)
\(\Rightarrow n_{N_2}=100.1=100\left(mol\right);n_{H_2}=500-100=400\left(mol\right)\)
ta có 200cm3=0,2 lítn hcl=2*0,2=0,4 molgọi số mol của caco3 là a,na2co3 là bcaco3 + 2hcl -> cacl2 + co2 + h2oa(mol)---2a(mol)--a-------a--------ana2co3 + 2hcl -> 2nacl + co2 + h2ob(mol)---2b(mol)---2b-------b------bta có100a+106b=20,62a+2b=0,4=> a=b=0,1 mol=> m caco3=10g; m na2co3=10,6 g
PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\) (1)
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\) (2)
a) Ta có: \(\Sigma n_{HCl}=0,2\cdot2=0,4\left(mol\right)\)
Gọi số mol của Na2CO3 là \(a\) \(\Rightarrow n_{HCl\left(1\right)}=2a\left(mol\right)\)
Gọi số mol của CaCO3 là \(b\) \(\Rightarrow n_{HCl\left(2\right)}=2b\left(mol\right)\)
Ta lập được hệ phương trình:
\(\left\{{}\begin{matrix}2a+2b=0,4\\106a+100b=20,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,1\cdot100=10\left(g\right)\\m_{Na_2CO_3}=10,6\left(g\right)\end{matrix}\right.\)
b) Theo PTHH: \(\left\{{}\begin{matrix}n_{NaCl}=2n_{Na_2CO_3}=0,2mol\\n_{CaCl_2}=n_{CaCO_3}=0,1mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\\m_{CaCl_2}=0,1\cdot111=11,1\left(g\right)\end{matrix}\right.\)
Mặt khác: \(\left\{{}\begin{matrix}m_{CO_2}=0,2\cdot44=8,8\left(g\right)\\m_{ddHCl}=200\cdot1,2=240\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{dd\left(saup/ư\right)}=m_{hh}+m_{ddHCl}-m_{CO_2}=251,8\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{11,7}{251,8}\cdot100\%\approx4,65\%\\C\%_{CaCl_2}=\dfrac{11,1}{251,8}\cdot100\%\approx4,41\%\end{matrix}\right.\)
\(n_{HCl} = \dfrac{448.1,12.3,65\%}{36,5} = 0,50176(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ ZnO + 2HCl \to ZnCl_2 + H_2O\\ n_{Zn} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)\\ n_{ZnO} = \dfrac{n_{HCl} - 2n_{Zn}}{2} = \dfrac{0,50176-0,1.2}{2} = 0,15088(mol)\\ \%m_{Zn} = \dfrac{0,1.65}{0,1.65 + 0,15088.81}.100\% = 34,72\%\\ \%m_{ZnO} = 65,28\%\)
Giả sử R có hóa trị n.
\(n_{HCl}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)
\(m_{ddH_2SO_4}=269,58.1,03=277,6674\left(g\right)\)
\(\Rightarrow m_{H_2SO_4}=277,6674.6\%\approx16,66\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{16,66}{98}=0,17\left(mol\right)\)
BTNT H, có: \(n_{H_2}=\dfrac{1}{2}n_{HCl}+n_{H_2SO_4}=0,21\left(mol\right)\)
Ta có: 24nMg + MR.nR = 3,96 (1)
BT e, có: 2nMg + n.nR = 0,21.2 (2) ⇒ 24nMg + 12n.nR = 5,04
⇒ 24nMg = 5,04 - 12n.nR
Thay vào (1) ta được MR.nR - 12n.nR = -1,08
\(\Rightarrow n_R=\dfrac{-1,08}{M_R-12n}\) \(\Rightarrow M_R-12n< 0\Rightarrow M_R< 12n\)
Với n = 1 ⇒ MR < 12 (loại)
Với n = 2 ⇒ MR < 24 (loại)
Với n = 3 ⇒ MR < 36. Mà MR > 24 ⇒ MR = 27 (g/mol)
→ R là Al.
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}24n_{Mg}+27n_{Al}=3,96\\2n_{Mg}+3n_{Al}=0,21.2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,03\left(mol\right)\\n_{Al}=0,12\left(mol\right)\end{matrix}\right.\)
Đkc hay đktc vậy nhìn nó hơi rối :v