a)

C% CuSO4 = 16/(16 + 184)  .100% = 8%
b)

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

CM NaOH = 0,5/4 = 0,125M

\(a.\)

\(m_{dd_{CuSO_4\:}}=16+184=200\left(g\right)\)

\(C\%_{CuSO_4}=\dfrac{16}{200}\cdot100\%=8\%\)

\(b.\)

\(n_{NaOH}=\dfrac{20}{40}=0.5\left(mol\right)\)

\(C_{M_{NaOH}}=\dfrac{0.5}{4}=0.125\left(M\right)\)

a)   \(m_{dmCUSO4}\) =  16+184 = 200g 

C% = \(\dfrac{16}{200}\)  x 100% =8 % 

b) \(n_{NaOH}\) = \(\dfrac{20}{40}\) = 0,5 (Mol)

\(C_M\) = \(\dfrac{0.5}{4}\) = 0,125 (M)  Vì 4000ml= 4l

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\

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

\(n_{H_2SO_4.4SO_3}=\dfrac{8,36}{418}=0,02\left(mol\right)\)

PTHH: H₂SO₄.4SO₃ + 4H₂O --> 5H₂SO₄

                  0,02------------------->0,1

\(\left\{{}\begin{matrix}n_{NaOH}=0,001V.1,5=0,0015V\left(mol\right)\\n_{KOH}=0,001V.0,5=0,0005V\left(mol\right)\end{matrix}\right.\)

PTHH: 2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

          0,0015V-->0,00075V

            2KOH + H₂SO₄ --> K₂SO₄ + 2H₂O

          0,0005V->0,00025V

=> 0,00075V + 0,00025V = 0,1

=> V = 100 (ml)

=> D

a, \(C\%_{KCl}=\dfrac{20}{20+60}.100\%=25\%\)

b, \(C\%=\dfrac{40}{40+150}.100\%\approx21,05\%\)

c, \(C\%_{NaOH}=\dfrac{60}{60+240}.100\%=20\%\)

d, \(C\%_{NaNO_3}=\dfrac{30}{30+90}.100\%=25\%\)

e, \(m_{NaCl}=150.60\%=90\left(g\right)\)

f, \(m_{ddA}=\dfrac{25}{10\%}=250\left(g\right)\)

g, \(n_{NaOH}=120.20\%=24\left(g\right)\)

Gọi: n_{NaOH (thêm vào)} = a (g)

\(\Rightarrow\dfrac{a+24}{a+120}.100\%=25\%\Rightarrow a=8\left(g\right)\)

a, \(n_{KOH}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)

b, \(n_{KOH\left(trong300mlA\right)}=0,3.0,3=0,09\left(mol\right)\)

Gọi: V_H2O = a (l)

\(\Rightarrow C_{M_A}=0,2=\dfrac{0,09}{a+0,3}\Rightarrow a=0,15\left(l\right)=150\left(ml\right)\)

Câu 1  :

a) n Na2O = 3,1/62 = 0,05(mol)

$Na_2O + H_2O \to 2NaOH$

Theo PTHH : n NaOH = 2n Na2O = 0,1(mol)

=> CM NaOH = 0,1/2 = 0,05M

Câu 2 : 

Coi n KOH = 1(mol)

=> V dd KOH = 1/2 = 0,5(lít) = 500(ml)

=> mdd KOH = D.V = 500.1,43 = 715(gam)

=> C% KOH = 1.56/715  .100% = 7,83%

1. Ta có : \(n_{Na_2O}=\dfrac{m}{M}=0,05mol\)

\(PTHH:Na_2O+H_2O\rightarrow2NaOH\)

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

\(\Rightarrow C_{MNaOH}=\dfrac{n}{V}=0,05M\)

2. - Gọi số lít KOH là a lít

\(\Rightarrow m_{dd}=D.V=1430a\left(g\right)\)

Mà \(n_{KOH}=C_M.V=2amol\)

\(\Rightarrow m_{KOH}=n.M=112a\left(g\right)\)

\(\Rightarrow C\%=\dfrac{m}{m_{dd}}.100\%=\dfrac{112a}{1430a}.100\%=~7,83\%\)

Đáp án C

Đáp án A

Vì sau phản ứng thu được hỗn hợp muối nên ta có NaOH hết