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\(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)
\(m_{H_2SO_4}=100.20\%=20\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{20}{98}=\dfrac{10}{49}\left(mol\right)\)
PTHH: CuO + H2SO4 → CuSO4 + H2O
Mol: 0,02 0,02 0,02
Ta có: \(\dfrac{0,02}{1}< \dfrac{\dfrac{10}{49}}{1}\) ⇒ CuO hết, H2SO4 dư
mdd sau pứ = 1,6 + 100 = 101,6 (g)
\(C\%_{ddCuSO_4}=\dfrac{0,02.160.100\%}{101,6}=3,15\%\)
\(C\%_{ddH_2SO_4}=\dfrac{\left(\dfrac{10}{49}-0,02\right).98.100\%}{101,6}=17,76\%\)
https://hoc24.vn/hoi-dap/question/280059.html Câu tương tự.
\(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.20\%}{98}=0,204\left(mol\right)\)
PTHH:
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
0,04 0,04 0,04
\(\dfrac{0,04}{1}< \dfrac{0,204}{1}\) --> H2SO4 dư
\(C\%_{CuSO_4}=\dfrac{0,04.160}{3,2+100}.100\%=6,2\%\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,2.98}{3,2+100}.100\%=19\%\)
ncuo= 1,6/80=0,02
nh2so4=(100*20)/( 98*100)= 0,2> 0,02-> cuo pư hết, h2so4 dư
cuo+ h2so4-> cuso4+h2o
0,02-> 0,02 0,02
mdd sau pư= 1,6+ 100= 101,6
c%h2so4 dư= (0,2-0,02)*98/101,6*100= 17,36%
c%cuso4= 0,02*160/101,6*100= 3,15%
nCuO= \(\frac{1,6}{80}\) = 0,02 (mol)
\(n_{H_2SO_4}\) = \(\frac{100.20\%}{98}\) =0,2041(mol)
CuO + H2SO4 \(\rightarrow\) CuSO4 + H2O
bđ 0,02 \(\frac{10}{49}\) (mol)
pư 0,02 \(\rightarrow\) 0,02 \(\rightarrow\) 0,02 (mol)
spư 0 0,1841 0,02 (mol)
md d (sau pư) = 100 + 1,6 = 101,6 (g)
C%(CuSO4) = \(\frac{0,02.160}{101,6}\) . 100% = 3,15%
C%(H2SO4)= \(\frac{0,1841.98}{101,6}\) . 100% = 17,76%
CuO + H2SO4 \(\rightarrow\)CuSO4 + H2O
nCuO=\(\dfrac{1,6}{80}=0,02\left(mol\right)\)
nH2SO4=\(\dfrac{100.20\%}{98}=0,2\left(mol\right)\)
Vì 0,2>0,02 nên H2SO4 dư 0,18(mol)
Theo PTHH ta có:
nCuO=nCuSO4=0,02(mol)
C% dd CuSO4=\(\dfrac{0,02.160}{1,6+100}.100\%=3,15\%\)
C% dd H2SO4=\(\dfrac{98.0,18}{101,6}.100\%=17,36\%\)
CuO+H2SO4--->CuSO4+H2O
b) n CuO=1,6/80=0,02(mol)
m H2SO4=100.20/100=20(g)
n H2SO4=20/98=0,2(mol)
--->H2SO4 dư
dd sau pư là H2SO4 dư và CuSO4
m dd sau pư=1,6+100=101,6(g)
n H2SO4=n CuO=0,02(mol)
m H2SO4=0,02.98=1,96(g)
m H2SO4 dư=20-1,96=18,04(g)
C% H2SO4 dư=18,04/101,6.100%=17,76%
n CuSO4=n CuO=0,02(mol)
m CuSO4=0,02.160=3,2(g)
C% CuSO4=3,2/101,6.100%=3,15%