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Ta có: \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
______0,5____0,5_____0,5_____0,5 (mol)
b, mH2SO4 = 0,5.98 = 49 (g)
c, mFeSO4 = 0,5.152 = 76 (g)
d, \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
____0,5__0,5 (mol)
⇒ mCu = 0,5.64 = 32 (g)
Bạn tham khảo nhé!
\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1 0,15 0 ,05 0,15
a)\(V=0,15\cdot22,4=3,36\left(l\right)\)
b)\(m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{14,7}{9,8}\cdot100=150\left(g\right)\)
c) \(m_{H_2}=0,15\cdot2=0,3\left(g\right)\)
\(m_{ddsau}=2,7+150-0,3=152,4\left(g\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\)
\(\Rightarrow C\%=\dfrac{17,1}{152,4}\cdot100=11,22\%\)
\(\left(a\right)2Al+3H_2O\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \left(b\right)n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ n_{Al}=\dfrac{0,6.2}{3}=0,4mol\\ m_{Al}=0,4.27=10,8g\\ \left(c\right)n_{O_2}=\dfrac{4,8}{32}=0,15mol\\ 4Al+3O_2\underrightarrow{t^0}2Al_2O_3\\ \Rightarrow\dfrac{0,4}{4}>\dfrac{0,15}{3}\Rightarrow Al.dư\\ n_{Al_2O_3}=\dfrac{0,15.2}{3}=0,1mol\\ m_{oxit}=m_{Al_2O_3}=0,1.102=10,2g\)
a: \(2Al+3H_2SO_4\rightarrow1Al_2\left(SO_4\right)_3+3H_2\uparrow\)
0,4 0,6 0,2 0,6
b: \(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)
=>\(n_{Al}=0.4\left(mol\right)\)
\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)
c: \(4Al+3O_2\rightarrow2Al_2O_3\)
0,4 0,2
\(m_{Al_2O_3}=0.2\left(27\cdot2+16\cdot3\right)=0.2\cdot102=20.4\left(g\right)\)
a) PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Hiện tượng: Chất bột tan dần, dd chuyển màu xanh
b+c) Ta có: \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{CuCl_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\\C_{M_{CuCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)
Số mol của đồng (II) oxit
nCuO = \(\dfrac{m_{CuO}}{M_{CuO}}=\dfrac{16}{80}=0,2\left(mol\right)\)
a) Pt : CuO + H2SO4 → CuSO4 + H2O\(|\)
1 1 1 1
0,2 0,2 0,2
Hiện tượng quan sát được : CuO bị hòa tan trong dung dịch H2SO4 tạo ra dung dịch có màu xanh lam
b) Số mol của dung dịch axit sunfuric
nH2SO4 = \(\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
Thể tích của dung dịch axit sunfuric cần dùng
CMH2SO4 = \(\dfrac{n}{V}\Rightarrow V=\dfrac{n}{C_M}=\dfrac{0,2}{2}=0,1\left(l\right)\)
c) Số mol cuả muối đồng (II) sunfat
nCuSO4 = \(\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
Nồng độ mol của của muối đồng (II) sunfat
CM = \(\dfrac{n}{V}=\dfrac{0,2}{0,1}=2\left(M\right)\)
Chúc bạn học tốt
a) PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
b+c)
Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)=n_{H_2SO_4}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{ddH_2SO_4}=\dfrac{0,3\cdot98}{30\%}=98\left(g\right)\end{matrix}\right.\)
d) PTHH: \(ZnSO_4+BaCl_2\rightarrow ZnCl_2+BaSO_4\downarrow\)
Ta có: \(n_{BaCl_2}=\dfrac{260\cdot20\%}{208}=0,25\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,25}{1}\) \(\Rightarrow\) ZnSO4 còn dư, BaCl2 phản ứng hết
\(\Rightarrow\left\{{}\begin{matrix}n_{ZnCl_2}=0,25mol=n_{BaSO_4}\\n_{ZnSO_4\left(dư\right)}=0,05mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,25\cdot136=34\left(g\right)\\m_{BaSO_4}=0,25\cdot233=58,25\left(g\right)\\m_{ZnSO_4\left(dư\right)}=0,05\cdot161=8,05\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{H_2}=0,3\cdot2=0,6\left(g\right)\)
\(\Rightarrow m_{dd}=m_{Zn}+m_{ddH_2SO_4}-m_{H_2}+m_{ddBaCl_2}-m_{BaSO_4}=318,65\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{34}{318,65}\cdot100\%\approx10,67\%\\C\%_{ZnSO_4\left(dư\right)}=\dfrac{8,05}{318,65}\cdot100\%\approx2,53\%\end{matrix}\right.\)
\(n_{CO_2\left(đktc\right)}=\dfrac{10,64}{22,4}=0,475\left(mol\right)\\a, K_2CO_3+2CH_3COOH\rightarrow2CH_3COOK+CO_2+H_2O\\ b,n_{K_2CO_3}=n_{CO_2}=0,475\left(mol\right)\\ \Rightarrow m_{K_2CO_3}=138.0,475=65,55\left(g\right)\\ n_{CH_3COOH}=0,475.2=0,95\left(mol\right)\\ C\%_{ddCH_3COOH}=\dfrac{0,95.60}{200}.100=28,5\%\)
a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)=n_{MgCl_2}\)
\(\Rightarrow m_{MgCl_2}=0,15\cdot95=14,25\left(g\right)\)
c) Theo PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(p.ứ\right)}=0,3\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl\left(p.ứ\right)}=0,3\cdot36,5=10,95\left(g\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=53,3\left(g\right)\)
\(\Rightarrow m_{ddHCl}=50\left(g\right)\) \(\Rightarrow C\%_{HCl\left(p.ứ\right)}=\dfrac{10,95}{50}\cdot100\%=21,9\%\)
PTHH: \(CaSO_3+H_2SO_4\rightarrow CaSO_4+H_2O+SO_2\uparrow\)
Ta có: \(n_{CaSO_3}=\dfrac{11,9}{120}\approx0,1\left(mol\right)=n_{CaSO_4}=n_{H_2SO_4}=n_{SO_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CaSO_4}=0,1\cdot136=13,6\left(g\right)\\m_{ddH_2SO_4}=\dfrac{0,1\cdot98}{5\%}=196\left(g\right)\\V_{SO_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)