a) Ta có: \(n_{Al\left(NO_3\right)_3}=\dfrac{4,26}{213}=0,02\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al^+}=0,02\left(mol\right)\\n_{NO_3^-}=0,06\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[Al^+\right]=\dfrac{0,02}{0,1}=0,2\left(M\right)\\\left[NO_3^-\right]=\dfrac{0,06}{0,1}=0,6\left(M\right)\end{matrix}\right.\)

b) Ta có: \(\left[Na^+\right]=0,1+0,02\cdot2+0,3=0,304\left(M\right)\)

c) Bạn xem lại đề !!

a) $n_{H_2SO_4}  = 0,35.2 = 0,7(mol)$
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$

Theo PTHH : 

$n_{NaOH} = 2n_{H_2SO_4} = 1,4(mol) \Rightarrow C_{M_{NaOH}} = \dfrac{0,7}{0,2} = 3,5M$

b) $V_{dd\ sau\ pư}  = 0,2 + 0,35 = 0,55(lít)$
$C_{M_{Na_2SO_4}} = \dfrac{0,7}{0,55} = 1,27M$

a) Ta có: \(n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,1}{0,5}=0,2\left(M\right)=\left[Na^+\right]=\left[Cl^-\right]\)

b) Ta có: \(n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\) 

\(\Rightarrow C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\) \(\Rightarrow\left\{{}\begin{matrix}\left[Ba^{2+}\right]=0,4\left(M\right)\\\left[OH^-\right]=0,8\left(M\right)\end{matrix}\right.\)

c) Ta có: \(n_{H_2SO_4}=0,025\cdot2=0,05\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,05}{0,125+0,025}\approx0,33\left(M\right)\) \(\Rightarrow\left\{{}\begin{matrix}\left[H^+\right]=0,66\left(M\right)\\\left[SO_4^{2-}\right]=0,33\left(M\right)\end{matrix}\right.\)

$n_{BaCl_2} = \dfrac{41,6}{208} = 0,2(mol)$
$C_{M_{BaCl_2}} = 0,2 : 0,5 = 0,4M$

$BaCl_2 \to Ba^{2+} + 2Cl^-$

$[Ba^{2+}] = 0,4M ; [Cl^-] = 0,4.2 = 0,8M$

ai giúp mình với ạ 

\(n_{NaOH}=\dfrac{200\cdot2\%}{40}=0.1\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{50\cdot49\%}{98}=0.25\left(mol\right)\)

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

\(0.1...............0.05...........0.05\)

\(n_{Na_2SO_4}=0.05\left(mol\right)\)

\(n_{H_2SO_4\left(dư\right)}=0.25-0.05=0.2\left(mol\right)\)

\(V_{dd}=\dfrac{200}{1}+\dfrac{50}{1.05}=247.6\left(ml\right)=0.2476\left(l\right)\)

\(\left[Na^+\right]=\dfrac{0.05\cdot2}{0.2476}=0.4\left(M\right)\)

\(\left[H^+\right]=\dfrac{0.2\cdot2}{0.2476}=1.6\left(M\right)\)

\(\left[SO_4^{2-}\right]=\dfrac{0.05+0.2}{0.2476}=1\left(M\right)\)

$n_{NaOH} = \dfrac{200.2\%}{40} = 0,1(mol)$
$n_{H_2SO_4} = \dfrac{50.49\%}{98} = 0,25(mol)$
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} : 2 < n_{H_2SO_4} : 1$ nên $H_2SO_4$ dư

$n_{H_2SO_4\ dư} = 0,25 - 0,1.0,5 = 0,2(mol)$

$n_{H^+\ dư} = 0,2.2 = 0,4(mol)$

Sau phản ứng :

$V_{dd} = \dfrac{200}{1} + 50.1,05 = 252,5(ml) = 0,2525(lít)$

Bảo toàn Na, S ta có : 

$[Na^+] = \dfrac{0,1}{0,2525} = 0,4M$
$[SO_4^{2-}] = \dfrac{0,25}{0,2525} = 0,99M$
$[H^+] = \dfrac{0,4}{0,2525} = 1,58M$

\(n_{OH^-}=\dfrac{200.2\%}{40}=0,1\left(mol\right);n_{H^+}=2.\dfrac{50.49\%}{98}=0,5\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

0,5.........0,1

=> Sau phản ứng H+ dư

\(n_{H^+\left(dư\right)}=0,5-0,1=0,4\left(mol\right)\)

Dung dịch sau phản ứng gồm các ion : Na ⁺ ,H⁺ dư, SO4^2-

V=\(\dfrac{200}{1}+\dfrac{50}{1,05}=247,6\left(ml\right)=0,2476\left(l\right)\)

\(\left[Na^+\right]=\dfrac{0,1}{0,2476}=0,4M\)

 \(\left[H^+_{dư}\right]=\dfrac{0,4}{0,2476}=0,8M\)

\(\left[SO_4^{2-}\right]=\dfrac{0,25}{0,2476}=1M\)

Đáp án A

a, \(n_{H^+}=n_{OH^-}=9.10^{-3}\left(mol\right)\Rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{\dfrac{9.10^{-3}}{2}}{0,05}=0,09M\)

b, \(\left[SO_4^{2-}\right]=\dfrac{4,5.10^{-3}}{0,05+0,15}=0,6M\)

\(\left[Na^+\right]=\dfrac{0,15.0,06}{0,05+0,15}=0,045M\)

\(\left[H^+\right]=\left[OH^-\right]=\dfrac{9.10^{-3}}{0,05+0,15}=0,045M\)