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a)
$CuO + H_2SO_4 \to CuSO_4 + H_2O$
Theo PTHH :
$n_{CuO} = n_{H_2SO_4} = \dfrac{98.40\%}{98} = 0,4(mol)$
$m = 0,4.80 = 32(gam)$
b)
$m_{dd\ sau\ pư} = 32 + 98 = 130(gam)$
$n_{CuSO_4} = n_{H_2SO_4} = 0,4(mol)$
$C\%_{CuSO_4} = \dfrac{0,4.160}{130}.100\% = 49,23\%$
a) \(m_{H_2SO_4}=98.40\%=39,2\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)
PTHH: CuO + H2SO4 → CuSO4 + H2O
Mol: 0,4 0,4 0,4
\(m_{CuO}=0,4.80=32\left(g\right)\)
b) mdd sau pứ = 32 + 98 = 130 (g)
\(C\%_{ddCuSO_4}=\dfrac{0,4.160.100\%}{130}=49,23\%\)
1)
a, \(n_{Al}=\dfrac{15,3}{102}=0,15\left(mol\right)\)
PTHH: Al2O3 + 6HCl → 2AlCl3 + 3H2O
Mol: 0,15 0,9 0,3
\(m_{ddHCl}=\dfrac{0,9.36,5.100}{20}=164,25\left(g\right)\)
b, mdd sau pứ = 15,3 + 164,25 = 179,55 (g)
c, \(C\%_{ddAlCl_3}=\dfrac{0,3.133,5.100\%}{179,55}=22,31\%\)
2)
a, \(m_{HCl}=54,75.20\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
PTHH: Al2O3 + 6HCl → 2AlCl3 + 3H2O
Mol: 0,05 0,3 0,1
\(m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)
b, mdd sau pứ = 5,1 + 54,75 = 59,85 (g)
\(C\%_{ddAlCl_3}=\dfrac{0,1.133,5.100\%}{59,85}=22,31\%\)
nAl2O3= 10,2/102= 0,1(mol)
a) PTHH: Al2O3 + 6 HCl -> 2 AlCl3 + 3 H2O
0,1_______0,6_______0,2_________0,3(mol)
mHCl=0,6.36,5= 21,9(g)
=>mddHCl= (21,9.100)/7,3=300(g)
b) mddsau= mAl2O3 + mddHCl= 10,2+300=310,2(g)
c) mAlCl3= 133,5.0,2=26,7(g)
=>C%ddAlCl3= (26,7/310,2).100=8,607%
\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)
PTHH: Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
______0,05------>0,15--------->0,05
=> mH2SO4 = 0,15.98 = 14,7(g)
=> \(C\%\left(H_2SO_4\right)=\dfrac{14,7}{100}.100\%=14,7\%\)
\(C\%\left(Fe_2\left(SO_4\right)_3\right)=\dfrac{0,05.400}{8+100}.100\%=18,52\%\)
PTHH: Fe2(SO4)3 + 6NaOH --> 2Fe(OH)3\(\downarrow\) + 3Na2SO4
________0,05----------------------->0,1
=> mFe(OH)3 = 0,1.107=10,7(g)
(Tiếp tục với bài làm của bạn @Thiên Quỳnh)
b) Ta có: \(n_{CuO}=n_{CuSO_4}=0,4mol\)
\(\Rightarrow m_{CuSO_4}=0,4\cdot160=64\left(g\right)\)
c) Ta có: \(m_{dd}=m_{Cu}+m_{ddH_2SO_4}=32+196=228\left(g\right)\)
\(\Rightarrow C\%_{CuSO_4}=\frac{64}{228}\cdot100\approx28,07\%\)
Fe+H2SO4->feSO4+H2
0,2--0,2---------0,2------0,2
n H2SO2=0,2 mol
=>m Fe=0,2.56=11,2g
=>Cm FeSO4=0,2\0,2=1M
\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\\ a.Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ 0,2........0,2.........0,2...........0,2\left(mol\right)\\ b.V_{dd.muối}=V_{ddH_2SO_4}=200\left(ml\right)=0,2\left(l\right)\\ C_{MddFeSO_4}=\dfrac{0,2}{0,2}=1\left(M\right)\)
\(m_{ct}=\dfrac{20.98}{100}=19,6\left(g\right)\)
\(n_{H2SO4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)
1 1 1 1
0,2 0,2 0,2
a) \(n_{CuO}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{CuO}=0,2.80=16\left(g\right)\)
b) \(n_{CuSO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{CuSO4}=0,2.160=32\left(g\right)\)
c) \(m_{ddspu}=16+98=114\left(g\right)\)
\(C_{CuSO4}=\dfrac{32.100}{114}=28,7\)0/0
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