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\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O|\)
1 2 1 1
0,1 0,2 0,1
b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
⇒ \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(C_{ddHCl}=\dfrac{7,3.100}{200}=3,65\)0/0
c) \(n_{CuCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{CuCl2}=0,1.135=13,5\left(g\right)\)
Chúc bạn học tốt
Ta có: \(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)
PTHH: MgO + 2HCl ---> MgCl2 + H2.
Theo PT: \(n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\)
=> \(m_{MgCl_2}=0,1.95=9,5\left(g\right)\)
Ta có: \(m_{dd_{MgCl_2}}=4+100=104\left(g\right)\)
=> \(C_{\%_{MgCl_2}}=\dfrac{9,5}{104}.100\%=9,13\%\)
Sửa đề cho dễ làm : dd K2CO3 13,8%
PTHH: \(2CH_3COOH+K_2CO_3\rightarrow2CH_3COOK+H_2O+CO_2\uparrow\)
a+b) Ta có: \(n_{CH_3COOH}=\dfrac{150\cdot12\%}{60}=0,3\left(mol\right)\)
\(\Rightarrow n_{K_2CO_3}=n_{CO_2}=0,15\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddK_2CO_3}=\dfrac{0,15\cdot138}{13,8\%}=150\left(g\right)\\V_{CO_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.\)
c) Theo PTHH:: \(n_{CH_3COOK}=0,3\left(mol\right)\) \(\Rightarrow m_{CH_3COOK}=0,3\cdot98=29,4\left(g\right)\)
Mặt khác: \(m_{CO_2}=0,15\cdot44=6,6\left(g\right)\)
\(\Rightarrow m_{dd}=m_{ddCH_3COOH}+m_{ddK_2CO_3}-m_{CO_2}=293,4\left(g\right)\)
\(\Rightarrow C\%_{CH_3COOK}=\dfrac{29,4}{293,4}\cdot100\%\approx10,02\%\)
Ta có: \(m_{Ba\left(OH\right)_2}=100.17,1\%=17,1\left(g\right)\Rightarrow n_{Ba\left(OH\right)_2}=\dfrac{17,1}{171}=0,1\left(mol\right)\)
\(m_{H_2SO_4}=150.9,8\%=14,7\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)
PT: \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_{4\downarrow}+2H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{BaSO_4}=n_{Ba\left(OH\right)_2}=0,1\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,15-0,1=0,05\left(mol\right)\)
Ta có: m dd sau pư = 100 + 150 - 0,1.233 = 226,7 (g)
\(\Rightarrow C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,05.98}{226,7}.100\%\approx2,16\%\)
a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
b, \(n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{0,3.98}{120}.100\%=24,5\%\)
c, m dd sau pư = 16,8 + 120 - 0,3.2 = 136,2 (g)
d, \(n_{FeSO_4}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow C\%_{FeSO_4}=\dfrac{0,3.152}{136,2}.100\%\approx33,48\%\)
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(n_{H_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\Rightarrow m_{Al}=0,4.27=10,8\left(g\right)\)
b, \(n_{HCl}=2n_{H_2}=1,2\left(mol\right)\Rightarrow C\%_{HCl}=\dfrac{1,2.36,5}{250}.100\%=17,52\%\)
c, m dd sau pư = 10,8 + 250 - 0,6.2 = 259,6 (g)
d, \(n_{AlCl_3}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow C\%_{AlCl_3}=\dfrac{0,4.133,5}{259,6}.100\%\approx20,57\%\)
\(n_{CuO}=\dfrac{1.6}{80}=0.02\left(mol\right)\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(0.02.......0.02.................0.02\)
\(m_{H_2SO_4}=0.02\cdot98=1.96\left(g\right)\)
\(m_{dd_{H_2SO_4}}=\dfrac{1.96}{20\%}=9.8\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng }}=1.6+9.8=11.4\left(g\right)\)
\(C\%_{CuSO_4}=\dfrac{0.02\cdot160}{11.4}=28.07\%\)
a) Đặt: nMg=x(mol); nZnO=y(mol)
nH2SO4= 0,2(mol)
PTHH: Mg + H2SO4 -> MgSO4 + H2
x___________x____x_______x(mol)
ZnO + H2SO4 -> ZnSO4 + H2O
y____y______y(mol)
Ta có:
\(\left\{{}\begin{matrix}24x+81y=12,9\\22,4x=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
mMg=0,2.24=4,8(g)
%mMg=(4,8/12,9).100=37,209%
=>%mZnO=62,791%
b) nH2SO4=x+y=0,3(mol)
=> \(C\%ddH2SO4=\dfrac{0,3.98}{120}.100=24,5\%\)
a) $CuO + H_2SO_4 \to CuSO_4 + H_2O$
Theo PTHH : $n_{H_2SO_4} = n_{CuO} = \dfrac{4}{80} = 0,05(mol)$
$m_{H_2SO_4} = 0,05.98 = 4,9(gam)$
b) $n_{CuSO_4} = n_{CuO} = 0,05(mol)$
$m_{dd\ sau\ pư} = m_{CuO} + m_{dd\ H_2SO_4} = 154(gam)$
$C\%_{CuSO_4} = \dfrac{0,05.160}{154}.100\% = 5,19\%$