\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

            0,1-------------->0,1---->0,05

=> V_H2 = 0,05.22,4 = 1,12 (l)

m_{dd sau pư} = 2,3 + 197,8 - 0,05.2 = 200 (g)

=> \(C\%=\dfrac{0,1.40}{200}.100\%=2\%\)

\(V_{dd}=\dfrac{200}{1,08}=\dfrac{5000}{27}\left(ml\right)=\dfrac{5}{27}\left(l\right)\)

=> \(C_M=\dfrac{0,1}{\dfrac{5}{27}}=0,54M\)

\(n_{Na}=\frac{4,6}{23}=0,2\left(mol\right)\) 

          \(2Na+2H_2O->2NaOH+H_2\)  (1)

theo (1) \(n_{NaOH}=n_{Na}=0,2\left(m0l\right)\) 

=> \(m_{NaOH}=0,2.40=8\left(g\right)\) 

theo (1) \(n_{H_2}=\frac{1}{2}n_{Na}=0,1\left(mol\right)\) 

khối lượng dung dịch sau phản ứng là 

  4,6 + 90 - 0,1.2=94,4(g)

nồng độ % dung dịch thu được là 

\(\frac{8}{94,4}.100\%\approx8,47\%\)

Na+H20-> NaOH + 1/2 H2

C%= mNaOH / mNa mH20 - mH2 nhân cho 100

\(n_{HCl}=\dfrac{44,8}{22,4}=2\)

\(\Rightarrow m_{HCl}=2.36,5=73g\)

=> \(C\%_{HCl}=\dfrac{73}{73+327}\times100\%=18,25\%\)

b. 

\(n_{HCl}=\dfrac{250.18,25\%}{36,5}=1,25mol\)

\(n_{CaCO_3}=\dfrac{50}{100}=0,5mol\)

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

\(n_{CaCl_2}=n_{CO_2}=0,5mol\)

\(n_{HClpu}=0,5.2=1mol\)

\(\Rightarrow n_{HCldu}=1,25-1=0,25\)

\(\Rightarrow m_{ddpu}=50+250-0,5.44=278g\)

\(C\%_{HCl}=\dfrac{0,25.36,5}{278}.100\%=3,28\%\)

\(C\%_{CaCl_2}=\dfrac{0,5.111}{278}.100\%=19,96\%\)

\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)

Pt : \(2Na+2H_2O\rightarrow2NaOH+H_2|\)

         2           2               2            1

       0,2                          0,2

\(n_{NaOH}=\dfrac{0,2.2}{2}=0,2\left(mol\right)\)

⇒ \(m_{NaOH}=0,2.40=8\left(g\right)\)

\(m_{ddspu}=2,3+100=102,3\left(g\right)\)

\(C_{ddNaOH}=\dfrac{8.100}{102,3}=7,82\)⁰/₀

 Chúc bạn học tốt

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(n_{H_2}=\dfrac{1}{2}n_{Na}=\dfrac{1}{2}.\dfrac{2,3}{23}=0,05\left(mol\right)\)

\(\Rightarrow m_{H_2}=0,05.2=0,1\left(g\right)\)

\(\Rightarrow m_{\text{dd sau pư}}=100+2,3-0,1=102,2\left(g\right)\)

\(n_{NaOH}=n_{Na}=0,1\left(mol\right)\)

\(\Rightarrow C\%=\dfrac{0,1.40}{102,2}.100\%=3,91\%\)

\(n_{K_2O}=\dfrac{11,28}{94}=0,12\left(mol\right)\)

PTHH :

\(K_2O+H_2O\rightarrow2KOH\)

0,12                    0,24 

\(m_{KOH}=0,24.56=13,44\left(g\right)\)

\(m_{ddKOH}=11,28+100=111,28\left(g\right)\)

\(C\%=\dfrac{13,44}{111,28}.100\%\approx12,08\%\)

\(a)2NaOH+CuCl_2\rightarrow Cu\left(OH\right)_2+2NaCl\\ b)n_{NaOH}=\dfrac{4}{40}=0,1mol\\ n_{CuCl_2}=n_{Cu\left(OH\right)_2}=0,1:2=0,05mol\\ m_{ddCuCl_2}=\dfrac{0,05.135}{10}\cdot100=67,5g\\ c)n_{NaCl}=n_{NaOH}=0,1mol\\ C_{\%NaCl}=\dfrac{0,1.58,5}{\dfrac{4}{10}\cdot100+67,5-0,05.98}\cdot100=14,0625\%\)

HCl k có phần trăm à

a) $2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
b)

n Al = 2,7/27 = 0,1(mol)

Theo PTHH :

n H2SO4 = 3/2 n Al = 0,15(mol)

=> C% H2SO4 = 0,15.98/441  .100% = 3,33%

c)

Theo PTHH :

n H2 = 3/2 n Al = 0,15(mol)

=> V H2 = 0,15.22,4 = 3,36(lít)