\(Zn+2HCl->ZnCl_2+H_2\\ Fe+2HCl->FeCl_2+H_2\\n_{Zn}=a;n_{Fe}=b\\ 65a+56b=2,1\\ 2\left(a+b\right)=0,4\\ a=-1,011;b=1,211\)

Đáp số ra số âm, không thoả mãn điều kiện thực tế a, b > 0.

`n_{Cl} = 2n_{MgCl_2} = 2.0,525 = 1,05(mol)`

Số nguyên tử `Cl` là:

`1,05.6.10^{23} = 6,3.10^{23}` (nguyên tử)

\(n_{nguyên.tử.Cl}=2n_{MgCl_2}=2.0,525=1,05\left(mol\right)\)

\(H_3PO_4+3NaOH->Na_3PO_4+3H_2O\\ V_{NaOH}=\dfrac{0,05\cdot0,5\cdot3}{1}=0,075L=75mL\)

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Ta có: \(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{HCl}=2n_{Fe}=0,5\left(mol\right)\\n_{H_2}=n_{Fe}=0,25\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow V_{H_2}=0,25.24,79=6,1975\left(l\right)\)

\(a=C_{M_{HCl}}=\dfrac{0,5}{0,1}=5\left(M\right)\)

b, Theo PT: \(n_{FeCl_2}=n_{Fe}=0,25\left(mol\right)\)

Ta có: \(n_{AgNO_3}=0,4.1,3=0,52\left(mol\right)\)

PT: \(2AgNO_3+FeCl_2\rightarrow Fe\left(NO_3\right)_2+2AgCl_{\downarrow}\)

______0,5______0,25______0,25________0,5 (mol)

\(AgNO_3+Fe\left(NO_3\right)_2\rightarrow Fe\left(NO_3\right)_3+Ag_{\downarrow}\)

0,02______0,02________0,02________0,02 (mol)

⇒ m = m_AgCl + m_Ag = 0,5.143,5 + 0,02.108 = 73,91 (g)

- Dd sau pư gồm: Fe(NO₃)₃: 0,02 (mol) và Fe(NO₃)₂: 0,25 - 0,02 = 0,23 (mol)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{Fe\left(NO_3\right)_3}}=\dfrac{0,02}{0,1+0,4}=0,04\left(M\right)\\C_{M_{Fe\left(NO_3\right)_2}}=\dfrac{0,23}{0,1+0,4}=0,46\left(M\right)\end{matrix}\right.\)

\(Fe+2HCl->FeCl_2+H_2\\ a.V=\dfrac{14}{56}\cdot22,4=5,6\left(L\right)\\ a=\dfrac{\dfrac{14}{56}\cdot2}{0,1}=5\left(M\right)\\ b.n_{AgNO_3}=0,4\cdot1,3=0,52mol\\ FeCl_2+AgNO_3->Fe\left(NO_3\right)_2+AgCl\\ Fe\left(NO_3\right)_2+AgNO_3->Ag+Fe\left(NO_3\right)_3\\ m=0,25\cdot143,5+0,25\cdot108=62,875\left(g\right)\\ C_{M\left(AgNO_3\right)}=\dfrac{0,02}{0,5}=0,04M\\ C_{M\left(Fe\left(NO_3\right)_3\right)}=\dfrac{0,25}{0,5}=0,5M\)

\(Al+\dfrac{3}{2}Cl_2->AlCl_3\\ Cu+Cl_2->CuCl_2\\ n_{Al}=a;n_{Cu}=b\\ 27a+64b=21,9\\ 1,5a+b=\dfrac{12,32}{22,4}=0,55\\ a=0,19;b=0,26\\ m_{AlCl_3}=133,5\cdot0,19=25,365g\\ m_{CuCl_2}=135\cdot0,26=35,1g\)

pứ có cần t^o không ạ?

\(Đặt.2.muối:ACO_3,B_2CO_3\\ n_{CO_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\\ PTHH:ACO_3+2HCl\rightarrow ACl_2+CO_2+H_2O\\ B_2CO_3+2HCl\rightarrow2BCl+CO_2+H_2O\\ n_{CO^{2-}_3}=n_{muối.cacbonat}=n_{CO_2}=0,3\left(mol\right)\\ n_{Cl^-}=2.0,3=0,6\left(mol\right)\\ m_{muối.khan}=m_{muối.cacbonat}+\left(m_{Cl^-}-m_{CO^{2-}_3}\right)=10+\left(35,5.0,6-60.0,3\right)=13,3\left(g\right)\)

cái phương trình hóa học của con B2CO3 em thấy có vấn đề ý ạ 

\(n_{Cl_2\left(giảm\right)}=n_{Cl_2\left(p.ứ\right)}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ Đặt:n_{Mg}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\Mg+Cl_2\rightarrow\left(t^o\right)MgCl_2\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ \Rightarrow\left\{{}\begin{matrix}24a+56b=13,6\\a+1,5b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\\ m_{Mg}=24a=2,4\left(g\right);m_{Fe}=56b=11,2\left(g\right)\)

\(2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{FeCl_3}=\dfrac{6,5}{162,5}=0,04\left(mol\right)\\ n_{Fe}=n_{FeCl_3}=0,04\left(mol\right)\Rightarrow m=m_{Fe}=0,04.56=2,24\left(g\right)\\ n_{Cl_2}=\dfrac{3}{2}.0,04=0,06\left(mol\right)\Rightarrow V=V_{Cl_2\left(đkc\right)}=0,06.24,79=1,4874\left(l\right)\)

\(\Delta_rH=-\dfrac{50,01kJ}{\dfrac{1}{26}mol}=-1300,26kJ\cdot mol^{-1}\)

Mặt khác, khi đốt 1 mol methane chỉ toả ra lượng nhiệt khoảng 890,2 kJ, thấp hơn nhiều so với acetylene nên acetylene được dùng trong đèn xì.