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\(n_{CuSO_4}=\dfrac{200.16\%}{160}=0,2\left(mol\right)\)
PTHH :
\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
0,2 0,4 0,2 0,2
\(m_{NaOH}=0,4.40=16\left(g\right)\)
\(m_{ddNaOH}=\dfrac{16.100}{10}=160\left(g\right)\)
\(c,m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)
\(m_{ddNa_2SO_4}=200+160-\left(0,2.98\right)=340,4\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{28,4}{240,4}.100\%\approx8,34\%\)
\(d,PTHH:\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
0,2 0,2
\(m_{CuO}=0,2.80=16\left(g\right)\)
a, \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
b, \(m_{CuSO_4}=200.16\%=32\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)
\(n_{NaOH}=2n_{CuSO_4}=0,4\left(mol\right)\Rightarrow m_{ddNaOH}=\dfrac{0,4.40}{10\%}=160\left(g\right)\)
c, \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,2\left(mol\right)\)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,2.142}{200+160-0,2.98}.100\%\approx8,34\%\)
d, \(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
\(n_{CuO}=n_{Cu\left(OH\right)_2}=0,2\left(mol\right)\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)
\(n_{CuSO_4}=\dfrac{200.16}{160.100}=0,2mol\)
\(n_{NaOH}=\dfrac{200.10}{40.100}=0,5mol\)
CuSO4+2NaOH\(\rightarrow\)Cu(OH)2\(\downarrow\)+Na2SO4
-Ta có tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\rightarrow\)CuSO4 hết, NaOH dư.
Cu(OH)2\(\overset{t^0}{\rightarrow}\)CuO+H2O
\(n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,2mol\)
a=\(m_{CuO}=0,2.80=16gam\)
\(m_{Cu\left(OH\right)_2}=0,2.98=19,6gam\)
\(n_{NaOH\left(pu\right)}=2n_{CuSO_4}=0,4mol\rightarrow n_{NaOH\left(dư\right)}=0,5-0,4=0,1mol\)
\(m_{NaOH\left(dư\right)}=0,1.40=4gam\)
\(n_{Na_2SO_4}=n_{CuSO_4}=0,2mol\rightarrow m_{Na_2SO_4}=0,2.136=27,2gam\)
\(m_{dd}=200+200-19,6=380,4gam\)
C%NaOH=\(\dfrac{4.100}{380,4}\approx1,05\%\)
C%Na2SO4=\(\dfrac{27,2.100}{380,4}\approx7,15\%\)
a. PTHH: 3NaOH + AlCl3 ---> Al(OH)3↓ + 3NaCl (1)
Ta có: \(C_{\%_{NaOH}}=\dfrac{m_{NaOH}}{100}.100\%=12\%\)
=> mNaOH = 12(g)
=> \(n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\)
Ta lại có: \(C_{\%_{AlCl_3}}=\dfrac{m_{AlCl_3}}{200}.100\%=13,35\%\)
=> \(m_{AlCl_3}=26,7\left(g\right)\)
=> \(n_{AlCl_3}=\dfrac{26,7}{133,5}=0,2\left(mol\right)\)
Ta thấy: \(\dfrac{0,3}{3}< \dfrac{0,2}{1}\)
Vậy AlCl3 dư
Theo PT(1): \(n_{Al\left(OH\right)_3}=\dfrac{1}{3}.n_{NaOH}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)
=> \(m_{Al\left(OH\right)_3}=0,1.78=7,8\left(g\right)\)
b. Ta có: \(m_{dd_{NaCl}}=12+200-7,8=204,2\left(g\right)\)
Theo PT(1): \(n_{NaCl}=n_{NaOH}=0,3\left(mol\right)\)
=> \(m_{NaCl}=0,3.58,5=17,55\left(g\right)\)
=> \(C_{\%_{NaCl}}=\dfrac{17,55}{204,2}.100\%=8,59\%\)
c. PTHH: 2Al(OH)3 ---to---> Al2O3 + 3H2O (2)
Theo PT(2): \(n_{Al_2O_3}=\dfrac{1}{2}.n_{Al\left(OH\right)_3}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
=> \(m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)
PTHH: \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
a) Ta có: \(n_{CuSO_4}=\dfrac{320\cdot20\%}{160}=0,4\left(mol\right)=n_{Cu\left(OH\right)_2}\) \(\Rightarrow m_{Cu\left(OH\right)_2}=0,4\cdot98=39,2\left(g\right)\)
b) Theo PTHH: \(n_{NaOH}=2n_{CuSO_4}=0,8mol\) \(\Rightarrow m_{ddNaOH}=\dfrac{0,8\cdot40}{10\%}=320\left(g\right)\)
c) Theo PTHH: \(n_{Na_2SO_4}=n_{CuSO_4}=0,4mol\) \(\Rightarrow m_{Na_2SO_4}=0,4\cdot142=56,8\left(g\right)\)
Mặt khác: \(m_{dd}=m_{ddNaOH}+m_{ddCuSO_4}-m_{Cu\left(OH\right)_2}=600,8\left(g\right)\)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{56,8}{600,8}\cdot100\%\approx9,45\%\)
Câu 1:
PTHH: 2Al + 3H2SO4 ===> Al2(SO4)3 + 3H2
a)Vì Cu không phản ứng với H2SO4 loãng nên 6,72 lít khí là sản phẩm của Al tác dụng với H2SO4
=> nH2 = 6,72 / 22,4 = 0,2 (mol)
=> nAl = 0,2 (mol)
=> mAl = 0,2 x 27 = 5,4 gam
=> mCu = 10 - 5,4 = 4,6 gam
b) nH2SO4 = nH2 = 0,3 mol
=> mH2SO4 = 0,3 x 98 = 29,4 gam
=> Khối lượng dung dịch H2SO4 20% cần dùng là:
mdung dịch H2SO4 20% = \(\frac{29,4.100}{20}=147\left(gam\right)\)
nH2 = 6.72 : 22.4 = 0.3 mol
Cu không tác dụng với H2SO4
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2
0.2 <- 0.3 <- 0.1 <- 0.3 ( mol )
mAl = 0.2 x 56 = 5.4 (g)
mCu = 10 - 5.4 = 4.6 (g )
mH2SO4 = 0.3 x 98 = 29.4 ( g)
mH2SO4 20% = ( 29.4 x100 ) : 20 = 147 (g)
\(n_{NaOH}=\frac{200.10}{100.40}=0,5\left(mol\right)\\ PTHH:2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ m_{ddCuSO_4}=\frac{\left(0,25.160\right).100}{16}250\left(g\right)\\ m_{kt}=98.0,25=24,5\left(g\right)\\ C\%_{ddNa_2SO_4}=\frac{0,25.142}{200+250}.100\%=7,89\left(\%\right)\)