a) n_KOH = 0,2.1 = 0,2 (mol)

PTHH: K₂O + H₂O --> 2KOH

              0,1<-----------0,2

=> m_K2O = 0,1.94 = 9,4 (g)

b)

m_KOH = 0,2.56 = 11,2 (g)

m_dd = 200.1,12 = 224 (g)

=> \(C\%=\dfrac{11,2}{224}.100\%=5\%\)

... câu a) tính m nào vậy ?

a) PTHH: \(SO_3+H_2O\rightarrow H_2SO_4\)

Ta có: \(n_{SO_3}=\dfrac{24}{80}=0,3\left(mol\right)=n_{H_2SO_4}\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,3\cdot98}{20\%}=147\left(g\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{147}{1,14}\approx128,95\left(ml\right)\)

b) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

Theo PTHH: \(n_{Fe}=n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)=n_{FeSO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,3\cdot56=16,8\left(g\right)\\V_{H_2}=0,3\cdot24,76=7,428\left(l\right)\\m_{FeSO_4}=0,3\cdot152=45,6\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}=163,2\left(g\right)\)

\(\Rightarrow C\%_{FeSO_4}=\dfrac{45,6}{163,2}\cdot100\%\approx27,94\%\)

a, \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

b, \(n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaO}=0,2\left(mol\right)\Rightarrow C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

c, \(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)

Theo PT: \(n_{H_2SO_4}=n_{Ca\left(OH\right)_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,2.98}{15\%}=\dfrac{392}{3}\left(g\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{\dfrac{392}{3}}{1,05}\approx124,44\left(ml\right)\)

giúp e vs mn ơi ngày mai e thi rồi\

\(a)CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ b)n_{Ca\left(OH\right)_2}=n_{CaO}=\dfrac{11,2}{56}=0,2mol\\ C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,2}{0,2}=1M\\ c)Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\\ n_{H_2SO_4}=n_{Ca\left(OH\right)_2}=0,2mol\\ V=V_{ddH_2SO_4}=\dfrac{0,2.98.100}{15}:1,05=124,4ml\)

\(n_{Na}=\dfrac{2.3}{23}=0.1\left(mol\right)\)

\(n_{HCl}=0.4\cdot2=0.8\left(mol\right)\)

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

\(0.8..............0.8\)

\(2Na+2H_2O\rightarrow2NaOH+H_{_{ }2}\)

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

\(n_{NaOH}=0.1< 0.8\)

Đề nhầm lãn !

Help meeee

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

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

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

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

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

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

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

a, \(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(n_{Na_2O}=\dfrac{3,1}{62}=0,05\left(mol\right)\)

Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,1\left(mol\right)\)

m _{dd sau pư} = 3,1 + 50 = 53,1 (g)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,1.40}{53,1}.100\%\approx7,53\%\)

b, \(2Na+2H_2O\rightarrow2NaOH+H_2\)

Ta có: \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{NaOH}=n_{Na}=0,2\left(mol\right)\\n_{H_2}=\dfrac{1}{2}n_{Na}=0,1\left(mol\right)\end{matrix}\right.\)

Ta có: m _{dd sau pư} = 4,6 + 95,6 - 0,1.2 = 100 (g)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,2.40}{100}.100\%=8\%\)