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\(a.Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{Zn}=n_{H_2}=0,2\left(mol\right)\\ \Rightarrow\%m_{Zn}=\dfrac{0,2.65}{25.8}.100=50,39\%\\ \%m_{Cu}=100-50,39=49,61\%\\ b.m_{ddsaupu}=0,2.65+200-0,2.2=212,6\left(g\right)\\ n_{ZnSO_4}=n_{H_2}=0,2\left(mol\right)\\ C\%_{ZnSO_4}=\dfrac{0,2.161}{212,6}.100=15,15\%\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{4,05}{27}=0,15\left(mol\right)\\n_{H_2SO_4}=\dfrac{294\cdot10\%}{98}=0,3\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,15}{2}< \dfrac{0,3}{3}\) \(\Rightarrow\) Axit còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,075\left(mol\right)=n_{H_2SO_4\left(dư\right)}\\n_{H_2}=0,225\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,225\cdot22,4=5,04\left(l\right)\\m_{Al_2\left(SO_4\right)_3}=0,075\cdot342=25,65\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,075\cdot98=7,35\left(g\right)\\m_{H_2}=0,225\cdot2=0,45\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Al}+m_{ddH_2SO_4}-m_{H_2}=297,6\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{25,65}{297,6}\cdot100\%\approx8,62\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{7,35}{297,6}\cdot100\%\approx4,47\%\end{matrix}\right.\)
\(n_{NaOH}=\dfrac{150.8\%}{40}=0,3\left(mol\right)\)
PTHH:
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,3 0,15 0,15 0,3
\(m_{HCl}=0,15.98=14,7\left(g\right)\)
\(m_{ddHCl}=\dfrac{14,7.100}{4,9}=300\left(g\right)\)
\(b,m_{Na_2SO_4}=0,15.142=21,3\left(g\right)\)
\(c,C\%_{Na_2SO_4}=\dfrac{21,3}{150+300}.100\%=4,733\%\)
Phần tính m dd axit bị nhầm thành thành HCl rồi em nhé, dẫn tới phần c cũng sai theo.
nCuO=16/80=0,2(mol)
a) PTHH: CuO + H2SO4 -> CuSO4 + H2O
0,2___________0,2_____0,2(mol)
b) mCuSO4=160.0,2=32(g)
c) mH2SO4=0,2.98=19,6(g)
=>C%ddH2SO4= (19,6/100).100=19,6%
Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{6}\left(mol\right)\Rightarrow m_{Al}=\dfrac{1}{6}.27=4,5\left(g\right)\)
b, \(n_{H_2SO_4}=n_{H_2}=0,25\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,25}{0,2}=1,25\left(M\right)\)
c, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=\dfrac{1}{12}\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=\dfrac{1}{12}.342=28,5\left(g\right)\)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ Mg+H_2SO_4\to MgSO_4+H_2\\ MgO+H_2SO_4\to MgSO_4+H_2O\\ \Rightarrow n_{Mg}=0,25(mol)\\ a,\begin{cases} \%_{Mg}=\dfrac{0,25.24}{14}.100\%=42,86\%\\ \%_{MgO}=100\%-42,86\%=57,14\% \end{cases}\\ b,n_{MgO}=\dfrac{14-0,25.24}{40}=0,2(mol)\\ \Rightarrow \Sigma n_{H_2SO_4}=0,2+0,25=0,45(mol)\\ \Rightarrow C\%_{H_2SO_4}=\dfrac{0,45.98}{200}.100\%=22,05\%\)
Sửa đề:
Cho 1,6g CuO tác dụng với 200g dụng dịch h2so4 a.Tính khối lượng h2so4 phản ứng b.tính nồng độ % của h2so4 c.tính khối lượng muối thu đc
Giải
\(a.n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\\ CuO+H_2SO_4\xrightarrow[]{}CuSO_4+H_2\\ n_{H_2SO_4}=n_{CuSO_4}=n_{CuO}=0,02mol\\ m_{H_2SO_4}=0,02.98=1,96\left(g\right)\\ b.m_{ddH_2SO_4}=1,6+200=201,6\left(g\right)\\ C_{\%H_2SO_4}=\dfrac{1,96}{201,6}\cdot100\%\approx0,97\%\\ c.m_{CuSO_4}=0,02.160=3,2\left(g\right)\)
Cuộc là sao ?
CuO hả?