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a, \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
b, \(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)
Theo PT: \(n_{CuO}=n_{CuCl_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,05.80=4\left(g\right)\)
c, \(C_{M_{CuCl_2}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
\(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)
PTHH :
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,05 0,1 0,05
\(b,m_{CuO}=0,05.80=4\left(g\right)\)
\(c,C_{M\left(CuCl_2\right)}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
nHCl=0,3.2=0,6(mol)
a) PTHH: CuO +2 HCl -> CuCl2 + H2O
0,3_______________0,6___0,3(mol)
b) mCuO=0,3.80=24(g)
c) VddCuCl2=VddHCl=0,3(l)
=>CMddCuCl2=0,3/0,3=1(M)
d) m(muối)=0,3.135=40,5(g)
\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O|\)
1 2 1 1
0,1 0,2 0,1
Hiện tượng : CuO tan dần , tạo ra dung dịch có màu xanh lam
b) \(n_{CuCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
⇒ \(m_{CuCl2}=0,1.135=13,5\left(g\right)\)
c) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(V_{ddHCl}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)
Chúc bạn học tốt
a,Hiện tượng: Sau phản ứng tạo thành dd màu xanh lam và có khi ko màu thoát ra
\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
PTHH: CuO + 2HCl → CuCl2 + H2O
Mol: 0,1 0,2 0,1
b, \(m_{CuCl_2}=0,1.135=13,5\left(g\right)\)
c, \(V_{ddHCl}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)
Bài 9 :
\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,05--->0,1-------->0,05
a) \(C_{MddHCl}=\dfrac{0,1}{0,1}=1\left(M\right)\)
b) \(m_{CuCl2}=0,05.135=6,75\left(g\right)\)
c) \(C_{MCuCl2}=\dfrac{0,05}{0,1}0,5\left(M\right)\)
Câu 10 :
\(n_{FeO}=\dfrac{3,6}{72}=0,05\left(mol\right)\)
\(FeO+2HCl\rightarrow FeCl_2+H_2O\)
0,05-->0,1------->0,05
\(m_{ddHCl}=\dfrac{0,1.36,5}{10\%}100\%=36,5\left(g\right)\)
\(m_{ddspu}=3,6+36,5=40,1\left(g\right)\)
\(C\%_{FeCl2}=\dfrac{0,05.127}{40,1}.100\%=15,84\%\)
\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\\a, CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4}=n_{H_2SO_4}=n_{CuO}=0,05\left(MOL\right)\\ b,m_{CuSO_4}=0,05.160=8\left(g\right)\\ c,V_{ddH_2SO_4}=\dfrac{0,05}{0,5}=0,1\left(l\right)\\ d,V_{ddCuSO_4}=V_{ddH_2SO_4}=0,1\left(l\right)\\ C_{MddCuSO_4}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)
Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
a, Theo PT: \(n_{CuCl_2}=n_{CuO}=0,1\left(mol\right)\Rightarrow m_{CuCl_2}=0,1.135=13,5\left(g\right)\)
b, \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
\(n_{FeO}=a\left(mol\right),n_{CuO}=b\left(mol\right)\)
\(m_{hh}=72a+80b=19.2\left(g\right)\left(1\right)\)
\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(n_{H_2SO_4}=a+b=0.25\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.15\)
\(m_{FeO}=0.1\cdot72=7.2\left(g\right)\)
\(m_{CuO}=12\left(g\right)\)
\(C_{M_{FeSO_4}}=\dfrac{0.1}{0.25}=0.4\left(M\right)\)
\(C_{M_{CuSO_4}}=\dfrac{0.15}{0.25}=0.6\left(M\right)\)
a) $CuO + 2HCl \to CuCl_2 + H_2O$
b) $n_{CuO} = \dfrac{1}{2}n_{HCl} = 0,025(mol)$
$m_{CuO} = 0,025.80 = 2(gam)$
c)
$n_{CuCl_2} = n_{CuO} = 0,025(mol)$
$C_{M_{CuCl_2}} = \dfrac{0,025}{0,05} = 0,5M$