\(n_K=\dfrac{31,2}{39}=0,8\left(mol\right)\)

PTHH :

\(2K+2H_2O\underrightarrow{t^o}2KOH+H_2\uparrow\)

0,8                    0,8       0,4 

\(a,V_{H_2}=0,4.22,4=8,96\left(l\right)\)

\(b,m_{KOH}=0,8.56=44,8\left(g\right)\)

\(m_{ddKOH}=\left(31,2+200\right)-\left(0,4.2\right)=300,4\left(g\right)\)

\(c,C\%_{KOH}=\dfrac{44,8}{\left(200+31,2\right)-\left(0,4.2\right)}.100\%\approx19,44\%\)

Câu 1 : 

\(n_{K_2O}=\dfrac{2.35}{94}=0.025\left(mol\right)\)

\(K_2O+H_2O\rightarrow2KOH\)

\(0.025...................0.05\)

\(C_{M_{KOH}}=\dfrac{0.05}{0.4}=0.125\left(M\right)\)

Câu 2 : 

\(n_{Ca\left(OH\right)_2}=\dfrac{1.11}{111}=0.01\left(mol\right)\)

\(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

\(0.01..........................0.01\)

\(C_{M_{CaCl_2}}=\dfrac{0.01}{0.5}=0.02\left(M\right)\)

Câu 3 : 

\(n_{Al_2\left(SO_4\right)_3}=0.6\cdot0.05=0.03\left(mol\right)\)

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

\(0.03...............................0.03\)

\(m_{Al_2O_3}=0.03\cdot102=3.06\left(g\right)\)

`a)m_[dd A]=[50,5.100]/10=505(g)`

`m_[H_2 O]=505-50,5=454,5(g)`

`b)n_[KNO_3]=[50,5]/101=0,5(mol)`

   `V_[dd B]=[0,5]/2=0,25(l)`

`c)20=[m+50,5]/[m+505].100`

`<=>m=63,125(g)`

a) 

\(n_{K_2O}=\dfrac{18,8}{94}=0,2\left(mol\right)\)

PTHH: K₂O + H₂O --> 2KOH

              0,2-->0,2---->0,4

m_ct = 0,4.56 = 22,4 (g)

m_dm = 81,2 - 0,2.18 = 77,6 (g)

m_dd = 22,4 + 77,6 = 100 (g)

b) 

\(C\%=\dfrac{22,4}{100}.100\%=22,4\%\)

c) 

\(C\%=\dfrac{22,4}{50+100}.100\%=14,933\%\)

d) 

\(m_{dd\left(sau.khi.thêm\right)}=\dfrac{22,4.100}{11,2}=200\left(g\right)\)

=> m_H2O(thêm) = 200 - 100 = 100 (g)

e) Gọi khối lượng KOH thêm là x (g)

Có: \(C\%_{\left(dd.sau.khi.thêm\right)}=\dfrac{22,4+x}{100+x}.100\%=30\%\)

=> x = 10,857 (g)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 2K + 2H₂O --> 2KOH + H₂

            0,3<-------------0,3<---0,15

=> m_K = 0,3.39 = 11,7 (g)

=> m_KOH(A) = 21,1 - 11,7 = 9,4 (g)

m_{KOH(dd sau pư)} = 0,3.56 + 9,4 = 26,2 (g)

a = 200 + 0,15.2 - 21,1 = 179,2 (g)

\(C\%=\dfrac{26,2}{200}.100\%=13,1\%\) => x = 13,1 

a) 

\(C\%_{dd.KOH}=\dfrac{7,5}{7,5+42,5}.100\%=15\%\)

b) \(n_{HNO_3}=\dfrac{1,26}{63}=0,02\left(mol\right)\Rightarrow C_{M\left(dd.HNO_3\right)}=\dfrac{0,02}{0,016}=1,25M\)

1)

$n_{Na_2O} = \dfrac{6,2}{62} = 0,1(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,2(mol)$
$m_{dd} = 6,2 + 193,8 = 200(gam) \Rightarrow C\%_{NaOH} = \dfrac{0,2.40}{200}.100\% = 4\%$

2) 

$n_{K_2O} = \dfrac{23,5}{94} = 0,25(mol)$
$K_2O + H_2O \to 2KOH$
$n_{KOH} = 2n_{K_2O} = 0,5(mol) \Rightarrow C_{M_{KOH}} = \dfrac{0,5}{0,5} = 1M$

3) $n_{Na_2O} = \dfrac{12,4}{62} = 0,2(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,4(mol)$
$C_{M_{NaOH}} = \dfrac{0,4}{0,5} =0,8M$

4)

$Na_2SO_3 + 2HCl \to 2NaCl +S O_2 + H_2O$
Theo PTHH : 

$n_{SO_2} = n_{Na_2SO_3} = \dfrac{12,6}{126} = 0,1(mol)$
$V_{SO_2} = 0,1.22,4 = 2,24(lít)$

5) $n_{CaO} = \dfrac{5,6}{56} = 0,1(mol)$

$CaO + 2HCl \to CaCl_2 + H_2O$

Theo PTHH : 

$n_{HCl} = 2n_{CaO} = 0,2(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{14,6\%} = 50(gam)$

Bài 1 : 

\(n_{Na2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

Pt : \(Na_2O+H_2O\rightarrow2NaOH|\)

           1          1              2

          0,1                      0,2

\(n_{NaOH}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

⇒ \(m_{NaOH}=0,2.40=8\left(g\right)\)

\(m_{ddspu}=6,2+193,8=200\left(g\right)\)

\(C_{ddNaOH}=\dfrac{8.100}{200}=4\)⁰/₀

 Chúc bạn học tốt

mH2O=71,8.1=71,8(g)

nK2O=28,2/94=0,3(mol)

PTHH: K2O + H2O -> 2 KOH

0,3____________0,6(mol)

mKOH= 0,6.56=33,6(g)

mddKOH= mK2O + mH2O= 28,2+71,8=100(g)

=>C%ddB=C%ddKOH=(33,6/100).100=33,6%

\(n_K=\frac{5,85}{15}=0,15(mol)\\ K+H_2O \to KOH +\frac{1}{2}H_2\\ n_{KOH}=n_K=0,15(mol)\\ n_{H_2}=\frac{1}{2}.n_K=\frac{1}{2}.0,15=0,075(mol)\\ m_{dd}=5,85+100-(0,075.2)=105,7(g)\\ C\%=\frac{0,15.56}{105,7}.100=7,95\%\)