\(C\%_{NaCl}=\dfrac{15}{15+45}.100\%=25\%\)

\(∑m_{dd}= m_{NaCl} + m_{H_2O} = 15 + 45 = 60(gam)\)

\(\text{C%}ddNaCl=\)\(\dfrac{15}{60}.100\%=25\%\)

\(C\%_{NaCl}=\dfrac{15}{15+45}.100\%=25\%\)

\(C\%_{KCl}=\dfrac{20}{600}\cdot100\%=3.33\%\)

\(C\%_{K_2SO_4}=\dfrac{75}{1500}\cdot100\%=5\%\)

\(C\%_{NaCl}=\dfrac{15}{15+45}\cdot100\%=25\%\)

\(n_{HCl}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(m_{HCl}=0.2\cdot36.5=7.3\left(g\right)\)

\(m_{dd_{HCl}}=7.3+500=507.3\left(g\right)\)

\(C\%_{HCl}=\dfrac{7.3}{507.3}\cdot100\%=1.44\%\)

a) \(C\%_{ddKCl}=\dfrac{20}{500}.100\%=4\%\)

b) \(m_{dd}=15+45=60\left(g\right)\)

\(C\%_{ddNaCl}=\dfrac{15}{60}.100\%=25\%\)

a)

C% KCl = 20/500  .100% = 4%

b)

m dd = m NaCl + m H2O = 15 + 45 = 60(gam)

C% NaCl = 15/60  .100% = 25%

a)C%KCl=20/500×100=4%

b)C%NaCl=15/(15+45)×100=25%

m dd = m NaCl + m H2O = 45 + 155 = 200(gam)

C% NaCl = 45/200 .100% = 22,5%

\(m_{dung\ dịch} =m_{NaCl} + m_{H_2O} = 45 + 155 = 200(gam)\\ C\%_{NaCl} = \dfrac{m_{NaCl}}{m_{dung\ dịch}}.100\% = \dfrac{45}{200}.100\% = 22,5\%\)

\(C\%=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{m_{ct}}{45+155}.100\%=\dfrac{45}{200}.100\%=22,5\%\)

\(m_{ct}=\dfrac{19,6.200}{100}=39,2\left(g\right)\)

\(n_{H2SO4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)

\(m_{ct}=\dfrac{5,2.200}{100}=10,4\left(g\right)\)

\(n_{BaCl2}=\dfrac{10,4}{208}=0,05\left(mol\right)\)

Pt : \(H_2SO_4+BaCl_2\rightarrow2HCl+BaSO_4|\)

            1             1             2             1

          0,4           0,05        0,1         0,05

a) Lập tỉ số so sánh: \(\dfrac{0,4}{1}>\dfrac{0,05}{1}\)

             ⇒ H₂SO₄ dư , BaCl₂ phản ứng hết

             ⇒ Tính toán dựa vào số mol của BaCl₂

\(n_{BaSO4}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

⇒ \(m_{BaSO4}=0,05.233=11,65\left(g\right)\)

b) \(n_{HCl}=\dfrac{0,05.2}{1}=0,1\left(mol\right)\)

⇒ \(m_{HCl}=0,1.36,5=3,65\left(g\right)\)

\(n_{H2SO4\left(dư\right)}=0,4-0,05=0,35\left(mol\right)\)

⇒ \(m_{H2SO4\left(dư\right)}=0,35.98=34,3\left(g\right)\)

\(m_{ddspu}=200+200-11,65=388,35\left(g\right)\)

\(C_{ddHCl}=\dfrac{3,65.100}{388,35}=0,94\)⁰/₀

\(C_{ddH2SO4\left(dư\right)}=\dfrac{34,3.100}{388,35}=8,83\)⁰/₀

 Chúc bạn học tốt

\(a.n_{H_2SO_4}=\dfrac{200.19,6\%}{98}=0,4\left(mol\right)\\ n_{BaCl_2}=\dfrac{200.5,2\%}{208}=0,05\left(mol\right)\\ BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ Vì:\dfrac{0,05}{1}< \dfrac{0,4}{1}\\ \Rightarrow H_2SO_4dư\\ n_{BaSO_4}=n_{BaCl_2}=0,05\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{BaSO_4}=0,05.233=11,65\left(g\right)\\ b.m_{ddsau}=200+200-11,65=388,35\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,05.2.36,5}{388,35}.100\approx0,94\%\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{\left(0,4-0,05\right).98}{388,35}.100\approx8,832\%\)

a)

Khối lượng của dung dịch:

\(m_{dd}=m_{ct}+m_{dm}=20+180=200\left(g\right)\)

Nồng độ phần trăm của dung dịch:

\(C\%=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{20}{200}.100\%=10\%\)

b) đề sai nha bạn