đề bài còn thiếu hay sai giá trị nào không bạn ?
a,
CuO + 2HCl ------> CuCl2 + H2O
0,2<---0,4--------->0,2------>0,2
b,
mCuCl2 = 0,2 . 135 = 27g
c,
CuO + H2SO4 ------> CuSO4 + H2O
0,2 --->0,2
m d2= 200g
\(n_{HCl}=C_M\cdot V=1\cdot0,4=0,4\left(mol\right)\)
PTHH :
CuO + 2HCl ----> CuCl2 + H2
.0,2........0,4.............0,2......0,2......(mol)
b) \(m_{CuCl_2}=x=n\cdot M=0,2\cdot135=27\left(g\right)\)
\(m_{CuO}=n\cdot M=0,2\cdot80=16\left(g\right)\rightarrow\%CuO=\dfrac{16}{20}\cdot100\%=80\%\)
\(\rightarrow\%m_{Cu}=100\%-80\%=20\%\)
C)
PTHH : CuO + H2SO4 ----> CuSO4 + H2O
.............0,2.........0,2.............0,2............0,2.....(mol)
\(\Rightarrow m_{H_2SO_4}=n\cdot M=0,2\cdot98=19,6\left(g\right)\)
\(\Rightarrow mdd_{H_2SO_4}=\dfrac{19,6\cdot100\%}{9,8\%}=200\left(g\right)\)
a) CuO + 2HCl → CuCl2 + H2↑ (1)
b) \(n_{HCl}=0,4\times1=0,4\left(mol\right)\)
Theo PT1: \(n_{CuO}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\times0,4=0,2\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,2\times80=16\left(g\right)\)
\(\Rightarrow x=m_{Cu}=20-16=4\left(g\right)\)
\(\%m_{CuO}=\dfrac{16}{20}\times100\%=80\%\)
\(\%m_{Cu}=\dfrac{4}{20}\times100\%=20\%\)
c) CuO + H2SO4 → CuSO4 + H2O (2)
Theo PT2: \(n_{H_2SO_4}=n_{CuO}=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,2\times98=19,6\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{19,6}{9,8\%}=200\left(g\right)\)
a) CuO + 2HCl -> CuCl2 + H2\(\uparrow\)(1)
b) nHCl = 0,4 . 1 = 0,4 mol
Theo phương trình 1 :
nCuO = \(\dfrac{1}{2}\) nHCl = \(\dfrac{1}{2}\) . 0,4 = 0,2 mol
-> mCuO = 0,2 . 80 = 16 g
-> x = mCu = 20 - 16 = 4 g
%mCuO = \(\dfrac{16}{20}.100\%=80\%\)
%mCu = \(\dfrac{4}{20}.100\%=20\%\)
c) CuO + H2SO4 -> CuSO4 + H2O (2)
Theo phương trình 2 :
nH2SO4 = nCuO = 0,2 mol
-> mH2SO4 =0,2 . 98 = 19,6 g
-> mddH2SO4 = \(\dfrac{19,6}{9,8\%}=200g\)