Có: \(n_{NaOH}=0,2.1=0,2\left(mol\right)\)

PT: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

______0,2_____0,1_______0,1 (mol)

a, \(V_{ddH_2SO_4}=\dfrac{0,1}{2}=0,05\left(l\right)\)

b, \(C_{M_{Na_2SO_4}}=\dfrac{0,1}{0,2+0,05}=0,4M\)

Bạn tham khảo nhé!

300ml = 0,3l

\(n_{HNO3}=1.0,3=0,3\left(mol\right)\)

Pt : \(NaOH+HNO_3\rightarrow NaNO_3+H_2O|\)

          1               1               1              1

         0,3            0,3            0,3

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

200ml = 0,2l

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

\(n_{NaNO3}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

⇒ \(m_{NaNO3}=0,3.85=25,5\left(g\right)\)

Sau phản ứng :

\(V_{dd}=0,2+0,3=0,5\left(l\right)\)

\(C_{M_{NaNO3}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\)

 Chúc bạn học tốt

\(n_{HNO_3}=0,3\left(mol\right)\)

\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)

Theo PT: \(n_{NaOH}=n_{NaNO_3}=n_{HNO_3}=0,3\left(mol\right)\)

\(\Rightarrow CM_{NaOH}=\dfrac{0,3}{0,2}=1,5M\)

\(m_{NaNO_3}=0,3.85=25,5\left(g\right)\)

\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)

PT: \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

Theo PT: \(n_{Na_2SO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)

\(n_{NaOH}=2n_{H_2SO_4}=0,4\left(mol\right)\)

\(\Rightarrow V_{ddNaOH}=\dfrac{0,4}{1}=0,4\left(l\right)\)

\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,2}{0,2+0,4}=\dfrac{1}{3}\left(M\right)\)

\(n_{HCl}=0,2.0,2=0,04\left(mol\right)\)

Để trung hòa thì:

\(n_{NaOH}=n_{HCl}\)

\(\Leftrightarrow0,1.V_{NaOH}=0,04\)

\(\Leftrightarrow V_{NaOH}=0,4\left(l\right)=400\left(ml\right)\)

PTHH: \(HCl+NaOH\rightarrow NaCl+H_2O\)

\(\Rightarrow n_{NaCl}=n_{HCl}=0,04\left(mol\right)\)

\(V_{dd}=0,2+0,4=0,6\left(l\right)\)

\(\Rightarrow C_{M\left(NaCl\right)}=\dfrac{0,04}{0,6}=0,67M\)

\(n_{HCl}=0,3.1=0,3\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaOH}=n_{NaCl}=n_{HCl}=0,3\left(mol\right)\\ V_{\text{dd}NaOH}=V=\dfrac{0,3}{2}=0,15\left(l\right)\\ C_{M\text{dd}A}=C_{M\text{dd}NaCl}=\dfrac{0,3}{0,15+0,3}=\dfrac{2}{3}\left(M\right)\)

PTHH : $NaOH + HCl \to NaCl + H_2O$
Theo PTHH : $n_{HCl} =  n_{NaOH} = 0,15.1 = 0,15(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,15}{0,2} = 0,75M$

Đáp án C

Câu 2 : 

$n_{HCl} = 0,2.1 + 0,3.1,5 = 0,65(mol)$
$V_{dd} = 0,2 + 0,3 = 0,5(mol)$
$C_{M_{HCl}} = \dfrac{0,65}{0,5} =1,3M$

Câu 3 : 

Gọi $m_{H_2O\ cần\ thêm} =a (gam)$

Sau khi thêm : 

$m_{NaOH} = 100.35\% = 35(gam)$
$m_{dd} = 100 + a(gam)$

Suy ra:  $\dfrac{35}{100 + a}.100\% = 20\%$

Suy ra: a = 75(gam)

Câu 4 : 

Gọi $V_{dd\ HCl\ 2M} =a (lít) ; V_{dd\ HCl\ 3M} = b(lít)$

Ta có : 

$a + b = 4$
$2a + 3b = 4.2,75$

Suy ra a = 1(lít) ; b = 3(lít)

nH2=\(\frac{6,72}{22,4}=0,3\)mol

PTHH

         M+2HCl--> MCl₂+H₂

      0,3mol<---------------0,3mol

=>M_M=\(\frac{19,5}{0,3}=64\)

=> km loại là kẽm (Zn)

b) n_NaOH=0,2.1=0,2 mol

PTHH

         NaOH+HCl-->NaCl + H2O

         0,2 mol--> 0,2 mol

---> thể tích HCl 1M đã dùng là V=\(\frac{0,2+0,3}{1}=0,5\)lít

=> C_M(ZnCl2)=\(\frac{0,3}{0,5}=0,6M\)