Mình đã giúp bạn ở dưới rồi nhé :)

`n_[Fe_2 O_3]=20/160=0,125(mol)`

`n_[HCl]=[10,95.300]/[100.36,5]=0,9(mol)`

`Fe_2 O_3 +6HCl->2FeCl_3 +3H_2 O`

  `0,125`        `0,75`       `0,25`                          `(mol)`

Ta có: `[0,125]/1 < [0,9]/6 =>HCl` dư, `Fe_2 O_3` hết

   `C%_[FeCl_3]=[0,25.162,5]/[20+300].100=12,7%`

   `C%_[HCl(dư)]=[(0,9-0,75).36,5]/[20+300].100=1,7%`

1

\(a)m_{H_2O}=250-5=245g\\b )C_{\%NaCl}=\dfrac{5}{250}\cdot100=2\%\)

\(2\\ m_{ddCuSO_4}=\dfrac{15.100}{5}=300g\\ m_{H_2O}=300-15=285g\)

Câu 1:

a, Ta có: m _dd = m _{chất tan} + m_H2O ⇒ m_H2O = 250 - 5 = 245 (g)

b, \(C\%_{NaCl}=\dfrac{5}{250}.100\%=2\%\)

Câu 2:

Ta có: \(C\%_{CuSO_4}=\dfrac{15}{m_{ddCuSO_4}}.100\%=5\%\)

\(\Rightarrow m_{ddCuSO_4}=300\left(g\right)\)

⇒ m_H2O = 300 - 15 = 285 (g)

a, \(C_{M_{H_2SO_4}}=\dfrac{0,1}{0,25}=0,4\left(M\right)\)

b, \(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,4}=1,25\left(M\right)\)

\(a)C_{M\left(H_2SO_4\right)}=\dfrac{0,1}{0,25}=0,4M\\ b)C_{M\left(NaOH\right)}=\dfrac{\dfrac{20}{40}}{0,4}=1,25M\)

\(a,n_{H_2SO_4}=0,1mol\)

\(V_{H_2SO_4}=0,25l\left(250ml=0,25l\right)\\ C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,1}{0,25}=0,4M\)

`b,` \(n_{NaOH}=\dfrac{m}{M}=\dfrac{20}{40}=0,5mol\)

\(V_{NaOH}=0,4l\left(400ml=0,4l\right)\)

\(C_{M_{NaOH}}=\dfrac{n}{V}=\dfrac{0,5}{0,4}=1,25M\)

1) ADCT:C%=mct/mdd×100%

→mct=C%×mdd÷100%

→mNaOH=10%×250÷100%=25(g)

2) ta có: nNaOH=20÷40=0,5(mol)

Đổi: 200ml=0,2l

ADCT:CM=n/V

→CMddNaOH=0,5÷0,2=2,5M

3) nNaOH=4÷40=0,1(mol)

ADCT: CM=n/V

→CM=0,1÷2=0,05M

1.

m_NaOH=10/100.250=25(g)

2.

n_NaOH=20/40=0,5(mol)

=>C_M=0,5/0,2=2,5M

Câu 1:

Khối lượng NaOH có trong 250g dd 10%:

\(m_{NaOH}=\dfrac{C\%dd_{NaOH}.m_{ddNaOH}}{100\%}=\dfrac{10.250}{100}=25\left(g\right)\)

Câu 2:

+) Ta có: \(\left[{}\begin{matrix}V_{ddNaOH}=200\left(ml\right)=0,2\left(l\right)\\n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\end{matrix}\right.\)

Nồng độ mol của dung dịch NaOH là:

\(C_{MddNaOH}=\dfrac{n_{NaOH}}{V_{ddNaOH}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)

Câu 3:

Ta có: \(\left[{}\begin{matrix}n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\\V_{ddNaOH}=2\left(l\right)\end{matrix}\right.\)

Nồng độ mol của dung dịch NaOH:

\(C_{MddNaOH}=\dfrac{n_{NaOH}}{V_{ddNaOH}}=\dfrac{0,1}{2}=0,05\left(M\right)\)

\(m_{dd}=20+250=270\left(g\right)\)

\(C\%_{muối}=\dfrac{20}{270}.100=7,407\%\)

mdd=20+250=270(g)

C%muối=20/270.100=7,407%

7,41%

Đáp án B

Độ tan : \(S = \dfrac{20,7}{100}.100 = 20,7(gam)\)

Công thức tính số tan : S = \(\dfrac{m_{chất\ tan}}{m_{dung\ môi}}\)

S(CuSO4,200C)= (20,7.100)/100=20,7(g)

=> Chọn B

S(CuSO4,200C)= (20,7.100)/100=20,7(g

\(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)

PTHH: 2K + Cl₂ --to--> 2KCl

            0,1   0,05               0,1

\(\rightarrow m_{KCl}=0,1.74,5=7,45\left(g\right)\\ V_{Cl_2}=0,05.22,4=1,12\left(l\right)\\ C\%_{KCl}=\dfrac{7,45}{250}=2,98\%\)

a, \(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)

2K + Cl₂ ----> 2KCl

0,1    0,05         0,1

\(V_{Cl_2}=0,05.22,4=1,12\left(l\right)\)

b, \(m_{KCl}=0,1.74,5=7,45\left(g\right)\)

\(\Rightarrow C\%_{ddKCl}=\dfrac{7,45.100\%}{250}=2,98\%\)

a) \(C\%_{ddKCl}=\dfrac{20}{500}.100\%=4\%\)

b) \(m_{dd}=15+45=60\left(g\right)\)

\(C\%_{ddNaCl}=\dfrac{15}{60}.100\%=25\%\)

a)

C% KCl = 20/500  .100% = 4%

b)

m dd = m NaCl + m H2O = 15 + 45 = 60(gam)

C% NaCl = 15/60  .100% = 25%