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

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

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

\(n_{SO2}=\dfrac{40}{64}=0,625\left(mol\right)\)

Pt : \(SO_2+H_2O\rightarrow H_2SO_3|\)

         1         1               1

      0,625                    0,625

\(n_{H2SO3}=\dfrac{0,625.1}{1}=0,625\left(mol\right)\)

250ml = 0,25l

\(C_{M_{H2SO3}}=\dfrac{0,625}{0,25}=2,5\left(M\right)\)

 Chúc bạn học tốt

a) \(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(C_{M_{ddCuSO_4}}=\dfrac{0,1}{0,2}=0,5M\)

b) \(m_{H_2SO_4}=\dfrac{150.14}{100}=21\left(g\right)\)

a)

$n_{CuSO_4} = \dfrac{16}{160} = 0,1(mol)$
$C_{M_{CuSO_4}} = \dfrac{0,1}{0,2} = 0,5M$

b)

$m_{H_2SO_4} = 150.14\% = 21(gam)$

0,1 ở đây có lẽ đơn vị là mol bạn nhỉ?

\(C_{M_{NaOH}}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)

a)

$Fe + H_2SO_4 \to FeSO_4 + H_2$
$n_{H_2SO_4} = n_{H_2} = n_{Fe} = \dfrac{16,8}{56} = 0,3(mol)$

$V = 0,3.22,4 = 6,72(lít)$
$C_{M_{H_2SO_4}} = \dfrac{0,3}{0,25} = 1,2M$

b)

$n_{CuO} = \dfrac{16}{80} = 0,2(mol)$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
$n_{CuO} < n_{H_2}$ nên $H_2$ dư

$n_{Cu} = n_{CuO} = 0,2(mol)$
$m_{Cu} = 0,2.64 = 12,8(gam)$

\(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ PT:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

       0,3     0,3                        0,3 (mol)

a) V= n. 22,4 = 0,3 . 22,4 = 6,72(l)

 \(C\%=\dfrac{m_{H_2SO_4}}{m_{dd}}.100\%=\dfrac{0,3.98}{200}.100\%=11,76\%\)

b) PT: \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)

         0,3                0,3 

=> m_Cu=n.M=0,3.64=19,2(g)

n Mg=4:24=1/6 mol

PTHH: Mg+H2SO4=> MgSO4+H2

             1/6-->1/6--------->1/6-->1/6

=> CM( H2SO4)=1/6:0,2=0,8M

Nmg=\(\frac{4}{24}\)=\(\frac{1}{6}\)(mol)

đổi 200ml=0,2(l)

pt: mg+h2so4->mgso4+h2

tlmol: 1 1 1 1

mol: \(\frac{1}{6}\) \(\frac{1}{6}\) \(\frac{1}{6}\) \(\frac{1}{6}\)

a) nồng độ mol của dung dịch h2so4 là

Cm=\(\frac{n}{V}\)=\(\frac{\frac{1}{6}}{0.2}\) =0,83M

câu b)của bn câu hỏi í là sao ??

a, PTPƯ: SO₃ + H₂O ---> H₂SO₄

n_SO3=\(\dfrac{2,24}{22,4}=0,1mol\)

1 mol SO₃ ---> 0,1 mol H₂SO₄

nên 0,1 mol SO₃ ---> 0,1 mol H₂SO₄

C_{M H2SO4}=\(\dfrac{0,1}{0,5}\)=0,2 M

b, PTPƯ: Zn + H₂SO₄ ---> ZnSO₄ + H₂

1 mol H₂SO₄ ---> 1 mol Zn

nên 0,1 mol H₂SO₄ ---> 0,1 mol Zn

m_Zn=0,1.65=6,5 g

a) 

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

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

\(n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)

=> \(C_M=\dfrac{0,2}{0,25}=0,8M\)

\(n_{CuSO_4}=\dfrac{32}{160}=0,2mol\)

250ml = 0,25l

\(C_{M_{CuSO_4}}=\dfrac{0,2}{0,25}=0,8M\)

`n_[CuSO_4] = 32 / 160 = 0,2 (mol)`

`C_[M_[CuSO_4]] = [ 0,2 ] / [ 0,25 ] = 0,8 (M)`