Bài 1:

\(m_{NaCl}=\frac{200.10}{100}=20\left(g\right)\)

Bài 2:

\(n_{HCl}=0,5.0,15=0,075\left(mol\right)\)

Câu 1:

\(n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\Rightarrow C_{M_{ddNaCl}}=\dfrac{0,1}{0,1}=1M\)

\(n_{KOH}=\dfrac{5,6}{56}=0,1\left(mol\right)\Rightarrow C_{M_{ddKOH}}=\dfrac{0,1}{0,1}=1M\)

\(n_{CaCl_2}=\dfrac{11,1}{111}=0,1\left(mol\right)\Rightarrow C_{M_{ddCaCl_2}}=\dfrac{0,1}{0,1}=1M\)

Câu 2:

a,\(n_{Ba\left(OH\right)_2}=0,2.1,5=0,3\left(mol\right)\)

b,\(C\%_{ddHCl}=\dfrac{0,2.36,5.100\%}{200}=3,65\%\)

  \(C\%_{ddH_2SO_4}=\dfrac{0,1.98.100\%}{200}=4,9\%\)

\(n_{HCl}=\frac{200.7,3}{100.36,5}=0,4\left(mol\right)\)

mHCl=\(\frac{200\cdot7,3\%}{100\%}\)=14,6(g)

nHCl=\(\frac{14,6}{36,5}\)=0,4(mol)

a)

m NaOH =200.10% =20(gam)

n NaOH = 20/40 = 0,5(mol)

b)

n H2SO4 = 0,2.0,25 = 0,05(mol)

m H2SO4 = 0,05.98 = 4,9(gam)

\(n_{NaOH}=\dfrac{200.4\%}{40}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{200.7,3\%}{36,5}=0,4\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ m_{ddsau}=200+200=400\left(g\right)\\ Vì:\dfrac{0,2}{1}< \dfrac{0,4}{1}\Rightarrow HCldư\\ n_{HCl\left(p.ứ\right)}=n_{NaCl}=n_{NaOH}=0,2\left(mol\right)\\ n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,2.36,5}{400}.100=1.825\%\\ C\%_{ddNaCl}=\dfrac{0,2.58,5}{400}.100=2,925\%\)

NaCl+HCl→NaCl+H2O

nHCl=300.7,3%/36,5=0,6>nNaOH=200.4%/40=0,2→HCl dư

nHCl pư=nNaCl=nNaOH=0,2(mol)

nHCl dư=0,6−0,2=0,4(mol)

mdd sau pư=300+200=500(gam)

C%NaCl=0,2.58,5500.100%=2,34%

C%HCl=0,4.36,5500.100%=2,92%

Trộn 200g dd chi đó bn

a) ta có : m_NaOH = 8g

b) ta có \(\left\{{}\begin{matrix}m_{ddH2SO4}=200g\\C\%=10\%\end{matrix}\right.\)

=> m_H2SO4 = 20 g

c)

a) \(n_{NaOH}=\frac{8}{40}=0,2\left(mol\right)\)

b) \(m_{H_2SO_4}=200\times10\%=20\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=\frac{20}{98}=\frac{10}{49}\left(mol\right)\)

c) \(n_{HCl}=0,3\times0,1=0,03\left(mol\right)\)

a) m_HCl = \(\frac{200.10\%}{100\%}\) = 20 g

b) 500 ml = 0,5 l

n_HCl = 3.0,5 = 1,5 mol

m_HCl = 1,5.36,5 = 54,75 g