Đáp án: C

giải chi tiết:

nKOH=11,2.20%56=0,04molnKOH=11,2.20%56=0,04mol

PTHH:2KOH+H2SO4→K2SO4+2H2O­­­­­­­­­­­­­0,04mol­­­­0,02molPTHH:2KOH+H2SO4→K2SO4+2H2O­­­­­­­­­­­­­0,04mol­­­­0,02mol

⇒

C

\(n_{KOH}=\dfrac{11,2.20\%}{56}=0.04mol\)

\(PTHH:2KOH+H_2SO_4->K_2SO_4+2H_2O\)

\(=>m_{H_2SO_4}=98.0,02=1,96\left(g\right)\)

\(=>m_{d^2H_2SO_4}35\%=\dfrac{1.96}{35\%}=5,6g\)

Đáp án: C

giải chi tiết:

\(n_{KOH}=\dfrac{11,2.20\%}{56}=0,04\left(mol\right)\)

PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

           0,04mol    0,02mol

\(\Rightarrow m_{H_2SO_4}=98.0,02=1,96\left(g\right)\)

\(\Rightarrow m_{d_2H_2SO_4}\)\(_{35\%}\) = \(\dfrac{1,96}{35\%}=5,6\left(g\right)\)

\(n_{KOH}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\)(mol)

PTHH : 2KOH + H₂SO₄ ---> K₂SO₄ + 2H₂O

                2           : 1          :     1        :   2

         0.2mol           0.1mol

\(m_{H_2SO_4}=n.M=0,1.98=9,8\left(g\right)\)

=> \(m_{ddH_2SO_4}=\dfrac{m_{H_2SO_4}.100\%}{C\%}=\dfrac{9,8.100\%}{35\%}=28\left(g\right)\)

\(n_{KOH}=\dfrac{11,2.20}{100.56}=0,04\left(mol\right)\)

PTHH: 2KOH + H₂SO₄ --> K₂SO₄ + 2H₂O

_____0,04---->0,02

=> m_H2SO4 = 0,02.98 = 1,96 (g)

=> \(m_{ddH_2SO_4}=\dfrac{1,96.100}{35}=5,6\left(g\right)\)

Hòa tan 23,5 gam K₂O vào nước. Sau đó dùng 250ml dung dịch HCl để trung hòa dung dịch trên. Tính nồng độ mol HCl cần dùng.

A. 1,5M                                 

B. 2,0 M                   

C. 2,5 M                                

D. 3,0 M.

$n_{K_2O} = \dfrac{23,5}{94} = 0,25(mol)$
$K_2O + H_2O \to 2KOH$
$HCl  + KOH \to KCl + H_2O$

Theo PTHH : $n_{HCl} = n_{KOH} = 2n_{K_2O} = 0,5(mol)$
$C_{M_{HCl}} = \dfrac{0,5}{0,25} = 2M$

Đáp án B

Ta có: \(C_{\%_{KOH}}=\dfrac{m_{KOH}}{11,2}.100\%=20\%\)

=> m_KOH = 2,24(g)

=> \(n_{KOH}=\dfrac{2,24}{56}=0,04\left(mol\right)\)

PTHH: 2KOH + H₂SO₄ ---> K₂SO₄ + 2H₂O

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}.n_{KOH}=\dfrac{1}{2}.0,04=0,02\left(mol\right)\)

=> \(m_{H_2SO_4}=0,02.98=1,96\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{1,96}{m_{dd_{H_2SO_4}}}.100\%=35\%\)

=> \(m_{dd_{H_2SO_4}}=5,6\left(g\right)\)

\(n_{OH^-}=n_{KOH}=\dfrac{1,12}{56}=0,02mol\)

Để trung hòa\(\Rightarrow n_{H^+}=n_{OH^-}=0,02mol\)

\(\Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{H^+}=0,01mol\Rightarrow m=0,98g\)

\(m_{dd}=\dfrac{0,98}{35}\cdot100=2,8\left(g\right)\)

Chọn C.

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

PTHH: Na₂O + H₂O ---> 2NaOH (1)

a. Theo PT₍₁₎: \(n_{NaOH}=2.n_{Na_2O}=2.0,05=0,1\left(mol\right)\)

=> \(C_{M_{NaOH}}=\dfrac{0,1}{2}=0,05M\)

b. PTHH: 2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}.n_{NaOH}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

=> \(m_{H_2SO_4}=0,05.98=4,9\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{4,9}{m_{dd_{H_2SO_4}}}.100\%=20\%\)

=> \(m_{dd_{H_2SO_4}}=24,5\left(g\right)\)

a.250ml=0,25l ; nHCl=0,25.1,5=0,375mol

KOH+HCl->KCl+H2O

1mol 1mol 1mol

0,375 0,375 0,375

VKOh=0,375/2=0,1875l

b.CM KCL=0,375/0,25=1,5M

c.NaOH+HCL=NaCl+H2O

1mol 1mol

0,375 0,375

mdd NaOH=0,375.40.100/10=150g

n_HCl = C_M . V = 1,5 . 0,25 = 0,375 mol

PTHH: KOH + HCl -------> KCl + H₂O

Pt: 1 1 1 1 (mol)

Pư 0,375 <-0,375 -----> 0,375-> 0,375 (mol)

a) V_KOH = \(\dfrac{n}{C_M}\)= 0,1875 l

b) C_M KCl = \(\dfrac{n}{V}\)= \(\dfrac{0,375}{0,25+0,1875}\)\(\approx\)0,86 M

c) PTHH: NaOH + HCL ----> NaCl + H₂O

Pt: 1 1 1 1 (mol)

Pư 0,375 <- 0,375 ------>0,375 --> 0,375 (mol)

m_ctNaOH = n . M = 15 g

m_{ddNaOH = \(\dfrac{15.100\%}{10\%}\)=} 150 g

:3

Câu 1 : 

\(n_{KOH}=\dfrac{25\%.112}{100\%.56}=0,5\left(mol\right)\)

Pt : \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

\(n_{H2SO4}=\dfrac{1}{2}n_{KOH}=0,25\left(mol\right)\Rightarrow m_{ddH2SO4}=\dfrac{0,25.98}{4,9\%}.100\%=500\left(g\right)\)

b,\(n_{HCl}=0,2.2=0,4\left(mol\right)\)

PTHH: CuO + 2HCl → CuCl₂ + H₂

Mol:     0,2         0,4  

\(\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)

c,\(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)

PTHH: ZnO + H₂SO₄ → ZnSO₄ + H₂

Mol:      0,2         0,2

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,2}{2}=0,1\left(l\right)\)

d,\(n_{H_2SO_4}=2.0,1=0,2\left(mol\right)\)

PTHH: H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

Mol:     0,2           0,4

\(\Rightarrow V_{ddKOH}=\dfrac{0,4}{1}=0,4\left(l\right)\)