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

PTHH: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

Theo PTHH: \(n_{H2}so_4=^nCuO=0,2mol\)

\(\rightarrow m_{H_2SO_4}=0,2.98=19,6g\)

\(+)^mddH_2SO_4=\dfrac{^mH_2SO_4}{C\%}.100=196g\)

Đến đây thì bn bt lm chx ạ?

\(m_{CH_3COOH}=12\%.100=12\left(g\right)\\ n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)

PTHH: CH₃COOH + NaOH ---> CH₃COONa + H₂O

               0,2--------->0,2------------>0,2

\(m_{NaOH}=0,2.40=8\left(g\right)\\ m_{ddNaOH}=\dfrac{8}{8,4\%}=\dfrac{2000}{21}\left(g\right)\\ m_{ddCH_3COONa}=\dfrac{2000}{21}+100=\dfrac{4100}{21}\left(g\right)\\ m_{CH_3COONa}=0,2.82=16,4\left(g\right)\\ C\%_{CH_3COONa}=\dfrac{16,4}{\dfrac{4100}{21}}.100\%=8,4\%\)

`=>` Gợi ý:

`CH3COOH + NaHCO3 => CH3COONa + CO2 + H2O`

`m<sub>CH3COOH</sub> = 100x12/100 = 12` (g)

`==> n<sub>CH3COOH</sub> = m/M = 12/60 = 0.2` (mol)

Theo pt: `=> n<sub>NaHCO3</sub> = 0.2` (mol)

`==> m<sub>NaHCO3</sub> = n.M = 0.2x84 =16.8` (g)

`==> m<sub>dd NaHCO3</sub> = 16.8x100/8.4 = 200` (g)

Ta có: `n<sub>CH3COONa</sub> = 0.2` (mol)

nNaOH = 4/40 = 0.1 (mol)

PTHH: NaOH + HCl -> NaCl + H2O

Từ PTHH: nNaCl = nHCl = nNaOH = 0.1 (mol)

a) mNaCl = 0.1*(23+35.5) = 5.85(g)

b) mHCl = 0.1*(1+35.5) = 3.65(g)

C%ddHCl = 3.65/100 * 100% = 3.65%

câu a là nồng độ mol hay nồng độ phần trăm

a)

$CuO + H_2SO_4 \to CuSO_4 + H_2O$

$n_{H_2SO_4} = n_{CuO} = \dfrac{1,6}{80} = 0,02(mol)$
$C\%_{H_2SO_4} = \dfrac{0,02.98}{100}.100\% = 1,96\%$

b)

$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} = 2n_{H_2SO_4} = 0,04(mol)$
$m_{NaOH} = 0,04.40 = 1,6(gam)$

c)

$Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O$

Cu dư nên $n_{SO_2} = \dfrac{1}{2}n_{H_2SO_4} = 0,05(mol)$
$V_{SO_2} = 0,05.22,4 = 1,12(lít)$

`Fe + 2HCl -> FeCl_2 + H_2`

`0,1`   `0,2`               `0,1`     `0,1`          `(mol)`

`n_[Fe]=[5,6]/56=0,1(mol)`

`a)m_[FeCl_2]=0,1.127=12,7(g)`

`b)C%_[HCl]=[0,2.36,5]/250 . 100=2,92%`

`c)C%_[FeCl_2]=[12,7]/[5,6+250-0,1.2].100~~4,97%`

200ml = 0,2l

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

        1         2              1           1

     0,3      0,6             0,3        0,3

a) \(n_{ZnCl2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(C_{M_{ZnCl2}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

b) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

c) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)

              1            1              1            1

            0,6          0,6

\(n_{NaOH}=\dfrac{0,6.1}{1}=0,6\left(mol\right)\)

\(m_{NaOH}=0,6.40=24\left(g\right)\)
\(m_{ddNaOH}=\dfrac{24.100}{20}=120\left(g\right)\)

 Chúc bạn học tốt

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

a) PTHH: CuO + H2SO4 -> CuSO4 + H2O

0,2___________0,2_____0,2(mol)

b) mCuSO4=160.0,2=32(g)

c) mH2SO4=0,2.98=19,6(g)

=>C%ddH2SO4= (19,6/100).100=19,6%