\(a)n_{H_2SO_4}=0,15.2=0,3mol\\ FeO+H_2SO_4\rightarrow FeSO_4+H_2O\\ n_{FeO}=n_{FeSO_4}=n_{H_2SO_4}=0,3mol\\ \%m_{FeO}=\dfrac{0,3.72}{22,4}\cdot100\%=96,43\%\\ \%m_{CuO}=100\%-96,43\%=3,57\%\\ b)C_{M_{FeSO_4}}=\dfrac{0,3}{0,15}=2M\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Na}=2.0,3=0,6\left(mol\right)\\ a,m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Na_2O}=26,2-13,8=12,4\left(g\right)\\b, n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ n_{NaOH\left(tổng\right)}=n_{Na}+2.n_{Na_2O}=0,6+\dfrac{12,4}{62}=0,8\left(mol\right)\\ m_{c.tan}=m_{NaOH}=0,8.40=32\left(g\right)\\ c,m_{ddNaOH}=m_{hh}+m_{H_2O}-m_{H_2}=26,2+200-0,3.2=225,6\left(g\right)\\ C\%_{ddNaOH}=\dfrac{32}{225,6}.100\approx14,185\%\)

Anh có làm rồi nha!

1.

2Na + 2H₂O\(\rightarrow\)2NaOH + H₂ (1)

Na₂O + H₂O \(\rightarrow\)2NaOH (2)

n_H2=\(\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PTHH 1 ta có:

2n_H2=n_Na=0,3(mol)

m_Na=0,3.23=6,9(g)

m_Na2O=13,1-6,9=6,2(g)

n_Na2O=\(\dfrac{6,2}{62}=0,1\left(mol\right)\)

Theo PTHH 1 và 2 ta có:

n_Na=n_NaOH=0,3(mol)

2n_Na2O=n_NaOH=0,2(mol)

m_NaOH=(0,2+0,3).40=20(g)

C% dd NaOH=\(\dfrac{20}{13,1+100-0,15.2}.100\%=17,73\%\)

2, gọi x la so mol cua Al

y la so mol cua Fe

2Al + 3H₂SO₄ \(\rightarrow\) Al₂(SO₄)₃ + 3H₂

de: x \(\rightarrow\) 1,5x \(\rightarrow\) 1,5x

Fe + H₂SO₄ \(\rightarrow\) FeSO₄ + H₂

de: y \(\rightarrow\) y \(\rightarrow\) y

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

Ta co: 27x + 56y = 22,2

1,5x + y = 0,6

=> x= 0,2 y= 0,3

b, \(\%m_{Al}=\dfrac{27.0,2}{22,2}.100\%\approx24,32\%\)

\(\%m_{Fe}=100-24,32\approx75,68\%\)

c, \(m_{H_2SO_4}=98.\left(0,2.1,5+0,3\right)=58,8g\)

\(C\%_{H_2SO_4}=\dfrac{58,8}{100}.100\%=58,8\%\)

C1:

nH2= 0,15(mol)

PTHH: Fe + 2 HCl-> FeCl2 + H2

0,15_____0,3______0,15___0,15(mol)

-> mFe= 0,15.56=8,4(g)

-> mCuO= 10-8,4=1,6(g)

b) -> nCuO= 0,02(mol)

PTHH: CuO +2 HCl -> CuCl2 + H2O

0,02_________0,04(mol)

c) nHCl (tổng)= 0,34(mol)

=> CMddHCl= 0,34/0,2=1,7(M)

C2:

a) Mg +2 HCl -> MgCl2 + H2

nH2= 0,2(mol) -> nMg= nMgCl2= nH2=0,2(mol); nHCl=0,4(mol)

mMg= 0,2.24=4,8(g) -> mCu= 5,2(g)

=> %mMg=(4,8/10).100=48%

=>%mCu= 100%-48%=52%

b) mMgCl2= 0,2.95=19(g)

mHCl=0,4.36,5=14,6(g) -> mddHCl= ?? Không cho C% sao tính ta

a) PTHH: Na2O + H20 -> 2NaOH
số mol Na20 = 0,25 (mol)
=> số mol NaOH = 0,5 mol.
Nôngd độ mol NaOH = 0,5 / 0,5 = 1 M
b) PTHH: H2SO4 + 2NaOH -> Na2SO4 + 2H2O
số mol H2SO4 = 1/2 số mol NaOH = 0,25 mol
C% H2SO4 = mH2SO4 / m ddH2SO4 . 100%
=> m ddH2SO4= 122,5 g
D=m/V => V= 107,5 ml

Số mol Na₂O = 15,5:62 = 0,25 mol

a) Khi cho Na₂O xảy ra phản ứng, tạo thành phản ứng dung dịch có chất tan là NaOH.

Na₂O + H₂O → 2NaOH

Phản ứng: 0,25 → 0,05 (mol)

500 ml = = 0,5 lít; C_{M, NaOH} = = 1M.

b) Phương trình phản ứng trung hòa dung dịch:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Phản ứng: 0, 5 → 0,25 0,25 (mol)

m_H2SO4 = 0,25x98 = 24,5 g

m_{dd H2SO4} = = 122,5 g

m_{dd, ml} = = ≈ 107,5 ml

a, PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

Gọi: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{FeO}=y\left(mol\right)\end{matrix}\right.\) ⇒ 80x + 72y = 11,2 (1)

Ta có: \(n_{H_2SO_4}=0,15.1=0,15\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{CuO}+n_{FeO}=x+y=0,15\left(2\right)\)

Từ (1) và (2) ⇒ x = 0,05 (mol), y = 0,1 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,05.80}{11,2}.100\%\approx35,71\%\\\%m_{FeO}\approx64,28\%\end{matrix}\right.\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{CuSO_4}=n_{Cu}=0,05\left(mol\right)\\n_{FeSO_4}=n_{FeO}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{CuSO_4}}=\dfrac{0,05}{0,15}=\dfrac{1}{3}\left(M\right)\\C_{M_{FeSO_4}}=\dfrac{0,1}{0,15}=\dfrac{2}{3}\left(M\right)\end{matrix}\right.\)