Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{Al2\left(SO4\right)3}=0,1.1=0,1\left(mol\right)\)
Pt : \(2K+2H_2O\rightarrow2KOH+H_2\)
0,3 0,3
\(6KOH+Al_2\left(SO_4\right)_3\rightarrow3K_2SO_4+2Al\left(OH\right)_3\)
0,3 0,1 0,15 0,1
a) A : khí H2 , D : Kết tủa Al(OH)3
b) Xét tỉ lệ : \(\dfrac{0,3}{6}< \dfrac{0,1}{1}=>Al_2\left(SO_4\right)_3dư\)
\(\Rightarrow m_D=m_{Al\left(OH\right)3}=0,1.78=7,8\left(g\right)\)
c) Dung dịch D gồm : Al2(SO4)3 dư và K2SO4
\(C_{MK2SO4}=\dfrac{0,15}{0,1}=1,5\left(M\right)\)
\(C_{MAl2\left(SO4\right)3dư}=\dfrac{0,1-\dfrac{0,3}{6}}{0,1}=0,5\left(M\right)\)
Chúc bạn học tốt
\(n_{CaCO_3}=0,05\left(mol\right)\)
\(n_{HCl}=\dfrac{100.3,65\%}{36,5.100\%}=0,1\left(mol\right)\)
\(CaCO_3+2HCl-->CaCl_2+H_2O+CO_2\uparrow\)
\(\dfrac{0,05}{1}=\dfrac{0,1}{2}\) => 2 chất hết
dd sau phản ứng CaCl2
\(C\%CaCl_2=\dfrac{0,1.36,5}{5+100-0,05.44}.100\%\approx3,55\%\)
Cảm ơn bạn nhiều nha mình đang cần gấp sau này có cái gì mong bạn giúp đỡ mình cảm ơn bạn lần nữa
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{Na}=n_{NaOH}=0,4:2=0,2mol\)
ta có \(\dfrac{n_{Na}}{n_{Al}}=\dfrac{1}{2}\)
\(\Rightarrow n_{Al}=2n_{Na}=2.0,2=0,4mol\\ m_{rắn}=m_{Al}=0,4.27=10,8g\)
\(C_M\) \(_A=C_M\) \(_{NaOH}=\dfrac{0,2}{0,4}=0,5M\)
\(Zn + Cl_2 \xrightarrow{t^o} ZnCl_2\\ n_{Zn} = \dfrac{13}{65} = 0,2 > n_{Cl_2} = 0,15\). Do đó.Zn dư
Zn + Cl2 \(\xrightarrow{t^o}\) ZnCl2
0,15.......0,15....0,15......................(mol)
Zn + 2AgNO3 → Zn(NO3)2 + 2Ag
0,05........................................... 0,1..........(mol)
ZnCl2 + 2AgNO3 → 2AgCl + Zn(NO3)2
0,15...............................0,3.........................(mol)
Vậy :
\(m_{kết\ tủa} = 0,1.108 + 0,3.143,5 = 53,85(gam)\)
\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)
\(Zn+Cl_2\underrightarrow{t^0}ZnCl_2\)
\(0.15....0.15...0.15\)
\(\Rightarrow Zndư,Cl_2hết\)
\(ZnCl_2+2AgNO3\rightarrow Zn\left(NO_3\right)_2+2AgCl\)
\(0.15..................................................0.3\)
\(m_{AgCl}=0.3\cdot143.5=430.05\left(g\right)\)
Chúc bạn học tốt !!!
a) Đặt \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow27a+24b=1,26\) (1)
Ta có: \(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)
Bảo toàn electron: \(3a+2b=0,12\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=n_{Al}=0,02\left(mol\right)\\b=n_{Mg}=0,03\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,02\cdot27}{1,26}\cdot100\%\approx42,86\%\\\%m_{Mg}=57,14\%\end{matrix}\right.\)
b) Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,02\left(mol\right)\\n_{MgCl_2}=n_{Mg}=0,03\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,02\cdot133,5=2,67\left(g\right)\\m_{MgCl_2}=0,03\cdot95=2,85\left(g\right)\end{matrix}\right.\)
Mặt khác: \(\left\{{}\begin{matrix}m_{ddHCl}=40\cdot1,25=50\left(g\right)\\m_{H_2}=0,06\cdot2=0,12\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{dd}=m_{KL}+m_{ddHCl}-m_{H_2}=51,14\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{2,67}{51,14}\cdot100\%\approx5,22\%\\C\%_{MgCl_2}=\dfrac{2,85}{51,14}\cdot100\%\approx5,57\%\end{matrix}\right.\)
\(n_{KMnO_4} = \dfrac{15,8}{158} = 0,1(mol)\\ n_{HCl} = 0,08.2 = 0,16(mol)\)
2KMnO4 + 16HCl \(\to\) 2KCl + 2MnCl2 + 5Cl2 + 8H2O
0,02.............0,16...................................0,05..............(mol)
\(n_{NaOH} = 0,2.1,5 = 0,3(mol)\)
2NaOH + Cl2 \(\to\) NaCl + NaClO + H2O
0,1..........0,05......0,05......0,05........................(mol)
Vậy :
\(C_{M_{NaCl}} = C_{M_{NaClO}} = \dfrac{0,05}{0,2}= 0,25M\\ C_{M_{NaOH}} = \dfrac{0,3-0,1}{0,2} = 1M\)
\(a)Bte:3n_{Fe}+3n_{Al}=n_{Ag}\\ \Leftrightarrow n_{Ag_3}=0,1.3+0,1.3\\ \Leftrightarrow n_{Ag}=0,6mol\\ m_{rắn}=m_{Ag}=0,6.108=64,8g\\ BTNT\left(Ag\right):n_{Ag}=n_{AgNO_3}=0,6mol\\ V_{AgNO_3}=\dfrac{0,6}{2}=0,3l\\ BTNT\left(Al\right):n_{Al}=2n_{Al_2O_3}\\ \Leftrightarrow0,1=2n_{Al_2O_3}\\ \Leftrightarrow n_{Al_2O_3}=0,05mol\\ BTNT\left(Fe\right):n_{Fe}=2n_{Fe_2O_3}\\ \Leftrightarrow0,1=2n_{Fe_2O_3}\\ \Leftrightarrow n_{Fe_2O_3}=0,05mol \\ b=m_{oxit.bazo}=0,05.\left(160+102\right)=13,1g\)
\(n_{HCl}=\dfrac{3,65\%.100}{100\%.36,5}=0,1\left(mol\right)\)
Pt : \(2Na+2HCl\rightarrow2NaCl+H_2\)
0,15 0,1 0,1 0,05
Xét tỉ lệ : \(\dfrac{0,15}{2}>\dfrac{0,1}{2}\Rightarrow Nadư\)
\(m_{ddspu}=0,15.23+100-0,05.2=103,35\left(g\right)\)
\(C\%_{NaCl}=\dfrac{0,1.58,5}{103,35}.100\%=5,66\%\)
Chúc bạn học tốt
\(n_{HCl}=\dfrac{100.3,65}{100}:3,65=0,1mol\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
0,15 0,15 0,075
\(NaOH+HCl\rightarrow NaCl+H_2O\\ \Rightarrow\dfrac{0,15}{1}>\dfrac{0,1}{1}\Rightarrow NaOH.dư\\ n_{HCl}=n_{NaOH}=n_{NaCl}=0,1mol\\ m_{dd}=0,15.23+100-0,075.2=103,3g\\ C_{\%NaCl}=\dfrac{0,1.58,5}{103,3}\cdot100=5,66\%\\ C_{\%NaOH\left(dư\right)}=\dfrac{\left(0,15-0,1\right).40}{103,3}\cdot100=1,94\%\)