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

\(a,n_{NaOH}=0,2.1=0,2\left(mol\right)\\ 2NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ n_{H_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MH_2SO_4}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\ b,n_{Na_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MNa_2SO_4}=\dfrac{0,1}{0,2+0,4}=\dfrac{1}{6}\left(M\right)\\ c,m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)

\(n_{CuO}=\dfrac{16}{80}=0,2mol\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

0,2        0,4          0,2         0,2

\(C_{M_{HCl}}=\dfrac{n_{HCl}}{V_{HCl}}=\dfrac{0,4}{0,2}=2M\)

\(m_{HCl}=0,4\cdot36,5=14,6g\)

nCuO = 16/80 = 0,2 (mol)

PTHH: CuO + 2HCl -> CuCl2 + H2

Mol: 0,2 ---> 0,4 ---> 0,2 ---> 0,2

CMCuCl2 = 0,2/0,2 = 1M

mHCl = 0,4 . 36,5 = 14,6 (g)

\(nCuO=\dfrac{80}{80}=1\left(mol\right)\)

\(CuO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)

1             2                            1                        1

\(m_{\left(muối\right)}=1.182=182\left(g\right)\)

\(mCH_3COOH=2.60=120\left(g\right)\)

sao có 100g dd axit mà tới 120g CH3COOH ta

Các bạn ơi giúp tôi dj mà đây là đề thj của trường tôi đó

Đây là hóa học mà

hsg hay đại trà

\(a/\\MgO+2HCl \to MgCl_2+H_2O\\ n_{MgO}=\frac{8}{40}=0,2(mol)\\ b/\\ n_{HCl}=0,2.2=0,4(mol)\\ CM_{HCl}=\frac{0,4}{0,2}=2M\)

\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)

a) Pt : \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

             1             2             1             1

           0,2            0,4

b) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

200ml =0,2l

\(C_{M_{ddHCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)

 Chúc bạn học tốt

$a\big)$

$Zn+2HCl\to ZnCl_2+H_2$

$CuO+H_2\xrightarrow{t^o}Cu+H_2O$

$b\big)$

$n_{Zn}=\dfrac{10,4}{65}=0,16(mol)$

Theo PT: $n_{Cu}=n_{Zn}=0,16(mol)$

$\to m_{Cu}=0,16.64=10,24(g)$

Ta có: \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Theo PT: \(n_{MgCl_2}=n_{H_2}=n_{Mg}=0,1\left(mol\right)\)

a, \(m_{MgCl_2}=0,1.95=9,5\left(g\right)\)

b, \(V_{H_2}=0,1.24,79=2,479\left(l\right)\)

c, \(n_{HCl}=2n_{Mg}=0,2\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{3,65\%}=200\left(g\right)\)

\(nCuO=\dfrac{80}{80}=1\left(mol\right)\)

\(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)

     2                     1                1                        1  (mol)

\(mCH_3COOH=2.60=120\left(g\right)\)

m muối = \(m\left(CH_3COO\right)_2Cu=1.182=182\left(g\right)\)

m H2O = 1.18 = 18 (g)

mdd = mddCH3COOH + m(CH3COO)2Cu + mH2O - mCuO

        = 100   + 182 + 18 - 80 = 220 (g)

\(C\%_{ddCH_3COOH}=\dfrac{120.100}{220}=54,55\%\)

a) \(n_{CuO}=\dfrac{80}{80}=1\left(mol\right)\)

PTHH: CuO + 2CH₃COOH ---> (CH₃COO)₂Cu + H₂O

              1---->2--------------------->1

=> m_muối = 1.182 = 182 (g)

b) \(C\%_{CH_3COOH}=\dfrac{60.2}{100}.100\%=120\%\) đề có sai không vậy bạn ?

\(m_{HCl}=50.7,3\%=3,65\left(g\right)\\ n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\\ m_{Zn}=0,05.65=3,25\left(g\right)\\ V_{H_2\left(ĐKTC\right)}=0,05.22,4=1,12\left(l\right)\\ m_{ZnCl_2}=0,05.136=6,8\left(g\right)\)

mHCl=50.7,3%=3,65(g) -> nHCl=0,1(mol)

a) PTHH: Zn + 2 HCl -> ZnCl2 + H2

nH2=nZnCl2=nZn=nHCl/2= 0,1/2=0,05(mol)

b) m=mZn=0,05.65=3,25(g)

c) V(H2,đktc)=0,05.22,4=1,12(l)

d) mZnCl2= 136.0,05= 7,8(g)