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Ta có: \(n_{CuO}=\dfrac{16}{80}=0,2mol\)
PTHH: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
Theo PTHH: \(n_{H2}so_4=^nCuO=0,2mol\)
\(\rightarrow m_{H_2SO_4}=0,2.98=19,6g\)
\(+)^mddH_2SO_4=\dfrac{^mH_2SO_4}{C\%}.100=196g\)
Đến đây thì bn bt lm chx ạ?
\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)
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)
\(m_{CH_3COOH}=12\%.100=12\left(g\right)\\ n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)
PTHH: CH3COOH + NaOH ---> CH3COONa + H2O
0,2--------->0,2------------>0,2
\(m_{NaOH}=0,2.40=8\left(g\right)\\ m_{ddNaOH}=\dfrac{8}{8,4\%}=\dfrac{2000}{21}\left(g\right)\\ m_{ddCH_3COONa}=\dfrac{2000}{21}+100=\dfrac{4100}{21}\left(g\right)\\ m_{CH_3COONa}=0,2.82=16,4\left(g\right)\\ C\%_{CH_3COONa}=\dfrac{16,4}{\dfrac{4100}{21}}.100\%=8,4\%\)
`=>` Gợi ý:
`CH3COOH + NaHCO3 => CH3COONa + CO2 + H2O`
`mCH3COOH = 100x12/100 = 12` (g)
`==> nCH3COOH = m/M = 12/60 = 0.2` (mol)
Theo pt: `=> nNaHCO3 = 0.2` (mol)
`==> mNaHCO3 = n.M = 0.2x84 =16.8` (g)
`==> mdd NaHCO3 = 16.8x100/8.4 = 200` (g)
Ta có: `nCH3COONa = 0.2` (mol)
a)
$CuO + H_2SO_4 \to CuSO_4 + H_2O$
$n_{H_2SO_4} = n_{CuO} = \dfrac{1,6}{80} = 0,02(mol)$
$C\%_{H_2SO_4} = \dfrac{0,02.98}{100}.100\% = 1,96\%$
b)
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} = 2n_{H_2SO_4} = 0,04(mol)$
$m_{NaOH} = 0,04.40 = 1,6(gam)$
c)
$Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O$
Cu dư nên $n_{SO_2} = \dfrac{1}{2}n_{H_2SO_4} = 0,05(mol)$
$V_{SO_2} = 0,05.22,4 = 1,12(lít)$
\(a,PTHH:CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2+2KCl\\ ...0,2......0,4.......0,2........0,4\left(mol\right)\\ b,n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\\ m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\\ c,m_{KOH}=0,4\cdot56=22,4\left(g\right)\\ m_{dd_{KOH}}=\dfrac{22,4\cdot100\%}{20\%}=112\left(g\right)\\ m_{dd_{KCl}}=m_{CuCl_2}+m_{dd_{KOH}}-m_{Cu\left(OH\right)_2}=27+112-19,6=119,4\left(g\right)\)
\(d,C\%_{dd_{KCl}}=\dfrac{74,5\cdot0,4}{119,4}\cdot100\%\approx24,96\%\)
a)PTHH: ZnCl2+2KOH---->Zn(OH)2+2KCl
b)
mZnCl2=204.10100=20,4(g)ZnCl2=204.10100=20,4(g)
nZnCl2=20,4136=0,15(mol)ZnCl2=20,4136=0,15(mol)
nKOH=112.20%56=0,4(mol)KOH=112.20%56=0,4(mol)
=> 0,15/1 < 0,4/1=> KOH dư
Theo pthh, ta có :
nCu(OH)2=nZnCl2=0,15(mol)Cu(OH)2=nZnCl2=0,15(mol)
mCu(OH)2=0,15.98=14,7(g)Cu(OH)2=0,15.98=14,7(g)
c) m dd sau pư=204+112=316(g)
Theo pthh
nKOH=2nZnCl2=0,3(mol)KOH=2nZnCl2=0,3(mol)
C% KOH=0,3.56326.100%=5,32%0,3.56326.100%=5,32%
nKCl=2nZnCl2=0,3(mol)KCl=2nZnCl2=0,3(mol)
C% KCl=0,3.74,5316.100%=7,07%
200ml = 0,2l
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,3 0,6 0,3 0,3
a) \(n_{ZnCl2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(C_{M_{ZnCl2}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
b) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)
c) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)
1 1 1 1
0,6 0,6
\(n_{NaOH}=\dfrac{0,6.1}{1}=0,6\left(mol\right)\)
\(m_{NaOH}=0,6.40=24\left(g\right)\)
\(m_{ddNaOH}=\dfrac{24.100}{20}=120\left(g\right)\)
Chúc bạn học tốt
a)nCuO= 16/80=0.2(mol)
CuO + H2SO4 ➞ CuSO4 +H2O
0.2........0.2.............0.2.........0.2...(mol)
mCuSO4=0.2*160=32(g)
mH2SO4=0.2*98=19.6(g)
mdung dịch H2SO4=19.6*100/10=196(g)
mdung dịch mới thu được=196+16=212(g)
C%CuSO4=32/212*100%=15.094%
b) CuSO4 + 2 NaOH ➞ Cu(OH)2 + Na2SO4
......0.2...............0.4.............0.2..............0.2.......(mol)
VNaOH= 0.4/2=0.2(l)
mCu(OH)2=0.2*98=19.6(g)
hình như không có phản ứng của axit sunfurơ với CuO bạn nhé