Ta có: \(n_{H_2}=\dfrac{0,5376}{22,4}=0,024\left(mol\right)\)

\(n_{OH^-}=2n_{H_2}=0,048\left(mol\right)\)

Gọi: \(\left\{{}\begin{matrix}n_{H_2SO_4}=x\left(mol\right)\\n_{HCl}=2x\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H^+}=2n_{H_2SO_4}+n_{HCl}=2x+2x=4x\left(mol\right)\)

PT: \(H^++OH^-\rightarrow H_2O\)

\(\Rightarrow4x=0,048\Rightarrow x=0,012\left(mol\right)\)

⇒ n_SO4^2- = n_H2SO4 = 0,012 (mol)

n_Cl^- = n_HCl = 0,024 (mol)

⇒ m _muối = m_KL + m_SO4^2- + m_Cl^- = 1,788 + 0,012.96 + 0,024.35,5 = 3,792 (g)

B

Đáp án D

Có nH₂SO₄ = x mol; n HCl = 2x mol

4x = 0,024.2 ⇒ x = 0,012 mol

⇒ m muối = 1,788 + 0,024.35,5 + 0,012.96 = 3,792 gam

Đáp án cần chọn là: B

Đáp án A

\(a.m_{HCl}=\dfrac{150.12}{100}=18\left(g\right)\\ n_{HCl}=\dfrac{18}{36,5}\approx0,5\left(mol\right)\\ b.m_{H_2SO_4}=\dfrac{50.10}{100}=5\left(g\right)\\ n_{H_2SO_4}=\dfrac{5}{98}\approx0,05\left(mol\right)\\ c.m_{AgNO_3}=\dfrac{100.12}{100}=12\left(g\right)\\ n_{AgNO_2}=\dfrac{12}{170}\approx0,07\left(mol\right)\)

a.

\(m_{HCl}=\dfrac{150.12\%}{100\%}=18\left(g\right)\\ \Rightarrow n_{HCl}=\dfrac{18}{36,5}\approx0,5\left(mol\right)\)

b. 

\(m_{H_2SO_4}=\dfrac{50.10\%}{100\%}=5\left(g\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{5}{98}\approx0,05\left(mol\right)\)

c. 

\(m_{AgNO_3}=\dfrac{100.12\%}{100\%}=12\left(g\right)\\ \Rightarrow n_{AgNO_3}=\dfrac{12}{170}\approx0,07\left(mol\right)\)

\(1,n_{H^+}=n_{Cl^-}=n_{HCl}=0,2\left(mol\right)\\2,n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,25\left(mol\right)\\ n_{OH^-}=2.0,25=0,5\left(mol\right)\\ 3,n_{H^+}=0,1.2=0,2\left(mol\right)\\ n_{SO_4^{2-}}=0,1\left(mol\right)\\ n_{Na^+}=n_{OH^-}=n_{NaOH}=0,15\left(mol\right) \)

Bài 1:

\(a.n_{NaOH\left(tổng\right)}=0,05.1+0,2.0,2=0,09\left(mol\right)\\ V_{ddNaOH\left(tổng\right)}=50+200=250\left(ml\right)=0,25\left(l\right)\\ C_{MddNaOH\left(cuối\right)}=\dfrac{0,09}{0,25}=0,36\left(M\right)\\ b.n_{HCl}=0,5.0,02=0,01\left(mol\right)\\ n_{H_2SO_4}=0,08.0,2=0,016\left(mol\right)\\ V_{ddsau}=20+80=100\left(ml\right)=0,1\left(l\right)\\ C_{MddH_2SO_4}=\dfrac{0,016}{0,1}=0,16\left(M\right)\\ C_{MddHCl}=\dfrac{0,01}{0,1}=0,1\left(M\right)\)

Bài 2:

\(a.m_{H_2SO_4}=29,4.10\%=2,94\left(g\right)\\ b.n_{H_2SO_4}=\dfrac{2,94}{98}=0,03\left(mol\right)\\ n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,01}{1}< \dfrac{0,03}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(dư\right)}=0,03-0,01=0,02\left(mol\right)\\ m_{H_2SO_4\left(dư\right)}=0,02.98=1,96\left(g\right)\\ n_{H_2}=n_{Fe}=0,01\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,01.22,4=0,224\left(l\right)\)