Hóa học

\(Na_2O+H_2SO_3\rightarrow Na_2SO_3+H_2O\\ Mg\left(OH\right)_2+K_2CO_3\rightarrow MgCO_3+2KOH\\ FeCl_3+Al\underrightarrow{t^o}Fe+AlCl_3\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ Ba\left(NO_3\right)_2+H_2SO_4\rightarrow BaSO_4+2HNO_3\)

n_KOH = 0,144 mol

Áp dụng định luật bảo toàn khối lượng ta có:

8,64 + 0,144.56 = 14,544 + m_nước

m_nước = 2,16 g ⇒ n_nước = 0,12 mol = n_axit

⇒ M_axit = 8,64 : 0,12 = 72 (g/mol)

Vậy axit là C2H3COOH (axit acrylic).

Câu 4: C

Câu 1: D

Câu 2: 2,3,1,3

Câu 3: C

Câu 5: B

a)

Gọi \(\left\{{}\begin{matrix}n_{HCl}=a\left(mol\right)\\n_{H_2SO_4}=2a\left(mol\right)\end{matrix}\right.\)

\(m_{NaOH}=100.10\%=10\left(g\right)\\ n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

PTHH:

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

a<--------a

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

a<----------2a

=> a + a = 0,25 

=> a = 0,125

=> \(\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,125.36,5}{100}.100\%=4,5625\%\\C\%_{H_2SO_4}=\dfrac{0,125.2.98}{100}.100\%=24,5\%\end{matrix}\right.\)

b)

PTHH:

\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)

0,25<-------0,25---------->0,25

\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

0,0625<---0,125------->0,0625

\(m_{dd.Ba\left(OH\right)_2}=\dfrac{\left(0,25+0,0625\right).171}{8,55\%}=625\left(g\right)\\ m_{dd.sau.pư}=625+100-0,25.233=666,75\left(g\right)\\ \Rightarrow C\%_{BaCl_2}=\dfrac{0,0625.208}{666,75}.100\%=1,95\%\)

\(n_{H_2SO_4}=0,5.0,3=0,15\left(mol\right)\\ pthh:Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
          0,05           0,15           0,05 
\(m_{Al_2O_3}=102.0,05=5,1\left(g\right)\\ m_{Al_2\left(SO_4\right)_3}=342.0,05=17,1\left(g\right)\\ n_{H_2SO_4}=n_{Al_2O_3}=0,15\left(mol\right)\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,15}{3}=0,05\left(l\right)hay50\left(ml\right)\)

\(n_{H_2SO_4}=0,5.0,3=0,15\left(mol\right)\\ pthh:Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)  
          0,05         0,15           0,05 
\(m_{Al_2O_3}=102.0,05=5,1\left(g\right)\\ m_{Al_2\left(SO_4\right)_3}=342.0,05=17,1\left(g\right)\) 
\(n_{HCl}=3n_{Al}=0,15\left(mol\right)\\ C_{M\left(HCl\right)}=\dfrac{0,15}{3}=0,05\left(l\right)hay50ml\)
          

a)

\(n_{AgCl}=\dfrac{14,35}{143,5}=0,1\left(mol\right)\)

PTHH: \(AgNO_3+HCl\rightarrow AgCl\downarrow+HNO_3\)

             0,1<------0,1<------0,1------->0,1

\(\left\{{}\begin{matrix}V_{dd.HCl}=\dfrac{0,1}{2}=0,05\left(l\right)\\V_{dd.AgNO_3}=\dfrac{0,1}{1}=0,1\left(l\right)\end{matrix}\right.\)

b) \(C_{M\left(HNO_3\right)}=\dfrac{0,1}{0,1+0,05}=\dfrac{2}{3}M\)

PTHH:

\(AgNO_3+HCl->AgCl+HNO_3\)

--> kết tủa là AgCl

\(n_{AgCl}=\dfrac{m}{M}=\dfrac{14,35}{108+35,5}=0,1\left(mol\right)\)

Dựa theo PTHH: 

--> \(n_{HCl}=n_{AgNO_3}=n_{HNO_3}=n_{AgCl}=0,1\left(mol\right)\)

\(V_{HCl}=\dfrac{n}{C_M}=\dfrac{0,1}{2}=0,05\left(lít\right)\)

\(V_{AgNO_3}=\dfrac{n}{C_M}=\dfrac{0,1}{1}=0,1\left(lít\right)\)

b) Dung dịch thu được sau phản ứng là \(HNO_3\)

\(V_{ddsau}=V_{HCl}+V_{AgNO_3}=0,05+0,1=0,15\left(lít\right)\)

\(C_{M\left(HNO_3\right)}=\dfrac{n}{V}=\dfrac{0,1}{0,15}=\dfrac{2}{3}M\)

\(m_{hh}=\dfrac{110g}{mol}\)

\(=>M_{O2}=\dfrac{110.43,46\%}{100}=48g\)

=>\(M_P=62\)g

\(m_{O2}=\dfrac{48}{16}=3mol\)

\(m_P=\dfrac{62}{31}=2mol\)

\(=>CTHH:P_2O_3\)

c2 : 
\(m_O=\dfrac{110.43,46}{100}=48\left(g\right)\\ m_P=110-48=62\left(g\right)\\ n_O=\dfrac{48}{16}=3\left(mol\right)\\ n_P=\dfrac{62}{31}=2\left(mol\right)\) 
-> CTHH là P₂O₃

hỏi 1 lần thôi bn =v

gọi cthh có dạng chung là P_xO_y (x,y>0) 
theo bài ra ta có 
\(\dfrac{31x}{56,54\%}=\dfrac{16y}{43,46\%}=\dfrac{110}{100\%}\) 
-> \(\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)(tm) 
-> CTHH : P₂O₃