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

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

a) Gọi số mol Mg, CuO là a, b (mol)

=> 24a + 80b = 14 (1)

\(n_{HCl}=\dfrac{255,5.10\%}{36,5}=0,7\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

             a--->2a--------->a------>a

           CuO + 2HCl --> CuCl₂ + H₂O

               b------>2b----->b

=> 2a + 2b = 0,7 (2)

(1)(2) => a = 0,25 (mol); b = 0,1 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,25.24}{14}.100\%=42,857\%\\\%m_{CuO}=\dfrac{0,1.80}{14}.100\%=57,143\%\end{matrix}\right.\)

b)

m_{dd sau pư} = 14 + 255,5 - 0,25.2 = 269 (g)

\(C\%_{MgCl_2}=\dfrac{0,25.95}{269}.100\%=8,829\%\)

\(C\%_{CuCl_2}=\dfrac{0,1.135}{269}.100\%=5,019\%\)

$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$

$n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)$

$\Rightarrow n_{Al}=0,15(mol)$

$\Rightarrow \%m_{Al}=\dfrac{0,15.27}{9,45}.100\%\approx 42,86\%$

$\Rightarrow \%m_{Cu}=100-42,86=57,14\%$

$b)$ Theo PT: $n_{HCl}=2n_{H_2}=0,45(mol)$

$\Rightarrow C_{M_{HCl}}=\dfrac{0,45.110\%}{0,5}=0,99M$

\(n_{CuO}=a\left(mol\right),n_{Fe_2O_3}=b\left(mol\right)\)

\(m=80a+160b=20\left(g\right)\left(1\right)\)

\(n_{HCl}=0.2\cdot3.5=0.7\left(mol\right)\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(n_{HCl}=2a+6b=0.7\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.05,b=0.1\)

\(m_{CuO}=0.05\cdot80=4\left(g\right)\)

\(m_{Fe_2O_3}=0.1\cdot160=16\left(g\right)\)

CuO + H2SO4---> CuSO4 +H2O

x                                   x mol

 Ag2O+ H2SO4 ----> Ag2SO4 +H2O

y                                       y mol

=> ta có hệ 80x+ 232y=54,4 và 160x+312y=78,4

=> x=0,1 và y=0,2 =>mCuO=8 gam và mAg2O=46,4 gam.

a) Na2CO3+2HCl-----> 2NaCl+H2O+CO2

n\(_{CO2}=\frac{0,448}{22,4}=0,02\left(mol\right)\)

Theo pthh

n\(_{Na2CO3}=n_{CO2}=0,02\left(mol\right)\)

m\(_{Na2CO3}=0,02.106=2,12\left(g\right)\)

m\(_{NaCl}=\)5-2,12=2,83(g)

b) mdd sau pư= 5+20-(0,02.44)=24,12(g)

theo pthh

n\(_{NaCl}=2n_{H2}=0,04\left(mol\right)\)

C%=\(\frac{0,04.58,5}{24,12}.100\%=9,7\%\)

Câu 2

a) 3NaOH+FeCl3---->Fe(OH)3+3NaCl

2Fe(OH)3---->Fe2O3+3H2O

b) Sai đề nhé.. tính C%ddA và m chất rắn B

m\(_{NaOH}=\)\(\frac{100.6}{100}=6\left(g\right)\)

n\(_{NaOH}=\frac{6}{40}=0,15\left(mol\right)\)

m\(_{FeCl3}=\)\(\frac{200.16,25}{100}=32,5\left(g\right)\)

n\(_{FeCl3}=\frac{32,5}{162,5}=0,2\left(mol\right)\)

=> FeCl3 dư

dd sau pư gồm NaCl và FeCl3

m dd sau pư= 100+200=300(g)

Theo pthh

n\(_{FeCl3}=\frac{1}{3}n_{NaOH}=0,05\left(mol\right)\)

n\(_{FeCl3}dư=0,2-0,05=0,15\left(mol\right)\)

C%FeCl3=\(\frac{0,15.162,5}{300}.100\%=8,125\%\)

Theo pthh1

n\(_{NaCl}=n_{NaOH}=0,15\left(mol\right)\)

C% NaCl =\(\frac{0,15.58,5}{300}.100\%=2,925\%\)

Chúc bạn học tốt

1) 2Na+2H2O → 2NaOH+H2

2) H2+CuOto→ Cu+H2O

Có nCuO=\(\dfrac{40}{80}\)=0,5 mol

Dựa vào PTHH 2) nH2=nCuO=0,5mol

Dựa vào PTHH 1) nNaOH=2nH2=0,5.2=1moll

Vậy mNaOH=1.40=40

→C%NaOH=\(\dfrac{40}{160}\).100%=25%

Bài 2:

Cho hỗn hợp 2 KL Na và Fe vào H₂O dư thì chỉ có Na pư

2Na + 2H₂O → 2NaOH + H₂

1 <----- 0,5

H₂ + CuO ---> Cu + H₂O

0,5 <-- 0,5

m_NaOH = 1 × 40 = 40 g

=> C% NaOH = m_ct/ m_dd ×100

= 40/160×100 = 25%