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\(CuO+H_2SO_{4\left(24,5\%\right)}\rightarrow CuSO_4+H_2O\)
\(Cu+2H_2SO_{4đ}\underrightarrow{t^o}CuSO_4+SO_2+2H_2O\)
\(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(\Rightarrow n_{Cu}=0,05\left(mol\right)\)
\(\Rightarrow m_{CuO}=10-64.0,05=6,8\left(g\right)\)
\(\Rightarrow n_{CuO}=0,085\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(24,5\%\right)}=0,085\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(24,5\%\right)}=8,33\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4\left(24,5\%\right)}=34\left(g\right)\)
a. PTHH:
\(Cu+H_2SO_4--\times-->\)
\(CuO+H_2SO_4--->CuSO_4+H_2O\left(1\right)\)
\(Cu+2H_2SO_{4_{đặc}}\overset{t^o}{--->}CuSO_4+SO_2+2H_2O\left(2\right)\)
Ta có: \(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Theo PT(2): \(n_{Cu}=n_{SO_2}=0,05\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,05.64=3,2\left(g\right)\)
\(\Rightarrow\%_{m_{Cu}}=\dfrac{3,2}{10}.100\%=32\%\)
\(\%_{m_{CuO}}=100\%-32\%=68\%\)
a)
$Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O$
$n_{Cu} = n_{SO_2} = \dfrac{1,12}{22,4} = 0,05(mol)$
$\%m_{Cu} = \dfrac{0,05.64}{10}.100\% = 32\%$
$\%m_{CuO} = 100\% -32\% = 68\%$
b)
$NaOH + SO_2 \to NaHSO_3$
$n_{NaOH} = n_{SO_2} = 0,05(mol)$
$V_{dd\ NaOH} = \dfrac{0,05}{2} = 0,025(lít) = 25(ml)$
\(A,\) Tên khí A là SO2
\(B,\) \(n_{SO_2}=\frac{1.12}{22.4}=0.05\left(mol\right)\)
\(K_2SO_3+H_2SO_4\rightarrow K_2SO_4+SO_2+H_2O\)
0.05 0.05 0.05
\(m_{H_2SO_4}=0.05\times98=4.9\left(g\right)\)
\(C\%_{H_2SO_4}=\frac{4.9}{24.5}\times100=20\%\)
\(C,\)
\(m_{K_2SO_3}=0.05\times158=7.9\left(g\right)\)
\(\%m_{K_2SO_3}=\frac{7.9}{21}\times100=37.6\%\)
\(\%m_{K_2SO_4}=100\%-37.6\%=62.4\%\)
PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(Cu+2H_2SO_{4\left(đ\right)}\underrightarrow{t^o}CuSO_4+SO_2+2H_2O\)
Ta có: \(n_{SO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Theo PT: \(n_{Cu}=n_{SO_2}=0,25\left(mol\right)\)
\(\Rightarrow\%m_{Cu}=\dfrac{0,25.64}{24}.100\%\approx66,67\%\)
a) $n_{Al} = 0,2(mol)$
b)
$n_{H_2SO_4} = \dfrac{294.20\%}{98} = 0,6(mol)$
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
$Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O$
$\Rightarrow n_{Al_2O_3} = \dfrac{0,6 - 0,2.1,5}{3} = 0,1(mol)$
$m = 0,1.102 = 10,2(gam)$
$n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = 0,2(mol)$
$m_{dd} = 0,2.27 + 10,2 + 294 - 0,3.2 = 309(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,2.342}{309}.100\% = 22,1\%$
Câu 1:a) nH2= 4,48/22,4=0,2(mol)
PTHH: Zn + 2 HCl -> ZnCl2 + H2
0,2_________0,4______0,2______0,2(mol)
nZn= nH2= 0,2(mol) => mZn= 0,2.65=13(g)
\(\%mZn=\frac{13}{21,1}.100\approx61,611\%\\ \Rightarrow\%mZnO\approx100\%-61,611\%\approx38,389\%\)
b) => mZnO= mhhA- mZn= 21,1-13=8,1(g)
=> nZnO= 8,1/81= 0,1(mol)
PTHH: ZnO + 2 HCl -> ZnCl2 + H2O
0,1_________0,2_____0,1(mol)
mZnCl2= 0,2.136+0,1.136=40,8(g)
mddsau= mhhA++mddHCl-mH2= 21,1+200 - 0,2.2= 220,7(g)
=> C%ddZnCl2= (40,8/220,7).100\(\approx18,487\%\)
