\(19.\\ a)n_{Mg}=\dfrac{2,4}{24}=0,1mol\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ n_{H_2}=n_{H_2SO_4}=n_{Mg}=0,1mol\\ V_{H_2}=0,1.22,4=2,24l\\ b)m_{ddH_2SO_4}=\dfrac{0,1.98}{10}\cdot100=98g\)

\(20.\\ a)n_{NaOH}=0,2.1=0,2mol\\ 2NaOH+H_2SO_4\rightarrow NaSO_4+2H_2O\)

0,2                0,1                0,1            

\(m_{Na_2SO_4}=0,1.142=14,2g\\ b)m_{ddH_2SO_4}=\dfrac{0,1.98}{5}\cdot100=196g\)

\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)

\(n_{HNO_3}=0,3.1=0,3\left(mol\right)\)

Theo PT : \(n_{NaOH}=n_{HNO_3}=0,3\left(mol\right)\)

=> \(CM_{NaOH}=\dfrac{0,3}{0,2}=1,5M\)

Theo PT : \(n_{NaNO_3}=n_{HNO_3}=0,3\left(mol\right)\)

\(CM_{NaNO_3}=\dfrac{0,3}{0,2+0,3}=0,6M\)

a) nHCl= (500/1000). 2= 1(mol)

nH2SO4= (500/1000).1= 0,5(mol)

PTHH: NaOH + HCl -> NaCl + H2O (1)

2NaOH + H2SO4 -> Na2SO4 + 2H2O (2)

Ta có: nNaOH = nNaOH (1) + nNaOH(2) = nHCl (1) + 2. nH2SO4 (2)= 1+ 2.0,5= 2(mol)

=> VddNaOH= 2/1= 2(M)

PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Ta có: \(n_{NaOH}=0,3\cdot0,1=0,03\left(mol\right)=n_{HCl\left(p/ứ\right)}\)

+) Trường hợp 1: NaOH còn dư

PTHH: \(Al+NaOH+H_2O\rightarrow NaAlO_2+\dfrac{3}{2}H_2\uparrow\)

Ta có: \(n_{NaOH\left(dư\right)}=n_{Al}=\dfrac{0,27}{27}=0,01\left(mol\right)\)

\(\Rightarrow n_{HCl}=n_{NaOH\left(p/ứ\right)}=0,02\left(mol\right)\) \(\Rightarrow a=C_{M_{HCl}}=\dfrac{0,02}{0,2}=0,1\left(M\right)\)

+) Trường hợp 2: HCl còn dư

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

Ta có: \(n_{Al}=0,01\left(mol\right)\) \(\Rightarrow n_{HCl\left(dư\right)}=0,03\left(mol\right)\)

\(\Rightarrow n_{HCl}=0,06\left(mol\right)\) \(\Rightarrow a=C_{M_{HCl}}=\dfrac{0,06}{0,2}=0,3\left(M\right)\) 

a)2Al+6HCl→2AlCl3+3H2

Fe+2HCl→FeCl2+H2

2Al+6H2SO4→Al2(SO4)3+3SO2+6H2O

2Fe+6H2SO4→Fe2(SO4)3+3SO2+6H2O

Cu+2H2SO4→CuSO4+SO2+2H2O

hh:Al(amol),Fe(bmol),Cu(cmol)

nNaOH=0,2×2=0,4mol

nHCl=0,4×2=0,8mol

⇒nHClpu=0,8−0,4=0,4mol

nSO2=5,6\22,4=0,25mol

27a+56b+64c=14,2

0,5a×3+0,5b×2=0,4

0,5a×1,5+0,5b×1,5+0,5c=0,25

⇒a=0,2;b=0,1;c=0,05

mAl=0,2×27=5,4g

mFe=0,1×56=5,6g

mCu=0,05×64=3,2g

b)mddspu=7,1+50−0,25×64=41,1g

C%Al2(SO4)3=41,6%

C%Fe2(SO4)3=24,33%

C%CuSO4=9,73%

Đề thiếu thông tin nhé

nOH-=nH+=0.4 mol
--->Cm NaOH= 0.4/0.2=2M...

Cái này a mk bảo nha... mk cx k hiểu lắm

mNaOH=25.4%=1g

=>nNaOH=1/40=0,025 mol

nH2SO4=0,2.52/1000=0,0104 mol

      2NaOH              +H2SO4=>Na2SO4        +2H2O

Bđ:0,025 mol

Pứ:0,0208 mol<=0,0104 mol=>0,0104 mol

Dư:4,2.10^(-3) mol

mNaOH dư=4,2.10^(-3).40=0,168g

mNa2SO4=0,0104.142=1,4768g

mdd sau pứ=25+51=76g

C%dd NaOH dư=0,168/76.100%=0,22%

C%dd Na2SO4=1,4768/76.100%=1,943%