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

a) 2NaOH + H₂SO₄ -- Na₂SO₄ + 2H₂O

b) \(n_{NaOH}=\dfrac{100.20}{100.40}=0,5\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ -- Na₂SO₄ + 2H₂O

______0,5----->0,25------>0,25

=> m_H2SO4 = 0,25.98 = 24,5 (g)

=> \(m_{ddH_2SO_4}=\dfrac{24,5.100}{19,6}=125\left(g\right)\)

c) m_Na2SO4 = 0,25.142 = 35,5 (g)

m_{dd sau pư} = 100 + 125 = 225 (g)

=> \(C\%\left(Na_2SO_4\right)=\dfrac{35,5}{225}.100\%=15,778\%\)

PTHH: \(2NaCl_{\left(r\right)}+H_2SO_{4\left(đ\right)}\underrightarrow{t^o}Na_2SO_4+2HCl\)

Ta có: \(n_{HCl}=\dfrac{50\cdot14,6\%}{36,5}=0,2\left(mol\right)=n_{NaCl}\)

\(\Rightarrow m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\)

Đáp án D

Chọn D

(H₂N)₂C₃H₅COOH

$n_{HCl} = 2n_X \to $ X có 2 nhóm $NH_2$

$n_{NaOH} = \dfrac{20.8\%}{40} = 0,04 = n_X \to$ X có 1 nhóm $COOH$

CTHH của X : $(NH_2)_2-R-COOH$

Muối là : 

$(NH_2)_2RCOONa  : 0,04(mol)$
$\Rightarrow 16.2 + R + 67 = \dfrac{5,6}{0,04} = 140$
$\Rightarrow R = 41(C_3H_5)$

Vậy X là : $(NH_2)_2C_3H_5COOH$