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CuO+H2SO4->CuSO4+H2O
0,02----------------0,02 mol
n CuO=1,6\80=0,02 mol
=>m CuSO4=0,02.160=3,2g
=>thiếu dữ kiện
\(n_{CuO}=\dfrac{1.6}{80}=0.02\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100\cdot20\%}{98}=\dfrac{10}{49}\left(mol\right)\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(TC:\dfrac{0.02}{1}< \dfrac{10}{49}\Rightarrow H_2SO_4dư\)
\(m_{dd}=1.6+100=101.6\left(g\right)\)
\(C\%_{CuSO_4}=\dfrac{0.02\cdot160}{101.6}\cdot100\%=3.15\%\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(\dfrac{10}{49}-0.02\right)\cdot98}{101.6}\cdot100\%=17.7\%\)
PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Ta có: \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,1\left(mol\right)\\n_{CuCl_2}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,1\cdot36,5}{7,3\%}=50\left(g\right)\\C\%_{CuCl_2}=\dfrac{0,05\cdot135}{4+50}\cdot100\%=12,5\%\end{matrix}\right.\)
Bài 1
\(a,n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ CuO+2HCl\xrightarrow[]{}CuCl_2+H_2O\\ n_{CuCl_2}=n_{CuO}=0,2mol\\ m_{CuCl_2}=0,2.135=27\left(g\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ C_{MHCl}=\dfrac{0,4}{0,5}=0,8\left(M\right)\)
Bài 5
\(a,n_{NaOH}=0,2.1=0,2\left(mol\right)\\ 2NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ n_{H_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MH_2SO_4}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\ b,n_{Na_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MNa_2SO_4}=\dfrac{0,1}{0,2+0,4}=\dfrac{1}{6}\left(M\right)\\ c,m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)
\(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.20\%}{98}=0,204\left(mol\right)\)
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%
\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
a) Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)
1 1 1 1
0,2 0,2 0,2
b) \(n_{H2SO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{H2SO4}=0,2.98=19,6\left(g\right)\)
\(C_{ddH2SO4}=\dfrac{19,6.100}{200}=9,8\)0/0
c) \(n_{CuSO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{CuSO4}=0,2.160=32\left(g\right)\)
\(m_{ddspu}=16+200=216\left(g\right)\)
\(C_{CuSO4}=\dfrac{32.100}{216}=14,81\)0/0
a) $Na_2O + 2HCl \to 2NaCl + H_2O$
b) $n_{Na_2O} = \dfrac{6,2}{62} = 0,1(mol)$
$n_{HCl} = 0,5(mol)$
Ta thấy :
$n_{Na_2O} : 1 < n_{HCl} : 2$ nên HCl dư
$n_{HCl\ pư} = 2n_{Na_2O} = 0,2(mol)$
$n_{HCl\ dư} = 0,5 - 0,2 = 0,3(mol)$
$C_{M_{HCl}} = \dfrac{0,3}{0,5} = 0,6M$
$C_{M_{NaCl}} = \dfrac{0,2}{0,5} = 0,4M$
\(CuO+H2SO4\Rightarrow CuSO4+H2O\)
H20 nha bạn