Hóa học

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\\ \Rightarrow m_{FeCl_2}=162,5.0,2=32,5\left(g\right)\\ \Rightarrow m=32,5\left(g\right)\)

a. PTHH: \(2H_2+O_2\rightarrow2H_2O\)

b. \(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)

Theo PTHH: \(n_{H_2}=n_{O_2}\cdot\dfrac{2}{1}=0,1\cdot\dfrac{2}{1}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c. Chưa hiểu đề bài lắm :))

d. Ta có: \(n_{O_2}=0,1\left(mol\right);n_{H_2}=\dfrac{3,6}{22,4}\approx0,16\left(mol\right)\)

Do \(0,1< 0,16\) nên \(H_2\) dư \(0,16-0,1=0,06\left(mol\right)\)

\(n_{KMnO_4}=1\left(mol\right)\\ a,2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\uparrow\\ a,n_{O_2}=\dfrac{n_{KMnO_4}}{2}=\dfrac{1}{2}=0,5\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,5.22,5=11,2\left(l\right)\\ \Rightarrow V_{kk}=5.V_{O_2\left(đktc\right)}=5.11,2=56\left(l\right)\\ b,2Mg+O_2\rightarrow\left(t^o\right)2MgO\\ n_{MgO}=2.n_{O_2}=2.0,5=1\left(mol\right)\\ \Rightarrow m_{MgO}=1.40=40\left(g\right)\)

a. PTHH: \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)

- Theo PTHH ta có:

\(n_O=n_{KMnO_4}\cdot\dfrac{1}{2}=1\cdot\dfrac{1}{2}=0,5\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,5.22,4=11,2\left(l\right)\)

- Do thể tích Oxi chiếm 1/5 thể tích không khí nên:

\(V_{KK}=11,2.5=56\left(l\right)\)

a) 

Rắn A: MgO và MgCO₃ dư( do rắn A tác dụng HCl có khí B bay ra nên MgCO₃ có dư)

Khí B: CO₂

ddC có dd Na₂CO₃ và dd NaHCO₃ (có NaHCO₃ vì dd C có tác dụng dd KOH)

b) Các PTHH:

\(MgCO_3\rightarrow\left(t^o\right)MgO+CO_2\uparrow\\ MgO+2HCl\rightarrow MgCl_2+H_2O\\ MgCO_3+2HCl\rightarrow MgCl_2+CO_2\uparrow+H_2O\\ CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ NaOH+CO_2\rightarrow NaHCO_3\\ Na_2CO_3+Ba\left(OH\right)_2\rightarrow BaCO_3\downarrow+2NaOH\\2KOH+2NaHCO_3\rightarrow K_2CO_3+Na_2CO_3+2H_2O\\ 2NaHCO_3+Ba\left(OH\right)_2\rightarrow BaCO_3\downarrow+Na_2CO_3+2H_2O\)

bn tk hoc247 hen:

a: \(n_{NaCl}=\dfrac{5.85}{58,5}=0.1\left(mol\right)\)

b: \(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)

\(n_X=\dfrac{P.V}{R.T}=\dfrac{1.1}{0,082.\left(273+134\right)}=0,03\left(mol\right)\\ M_X=29.1,5862=46\left(\dfrac{g}{mol}\right)\\ Đặt.CTTQ:N_kO_t\left(k,t:nguyên,dương\right)\\ Ta.có:k:t=\dfrac{30,43\%}{14}:\dfrac{69,57\%}{16}=0,022:0,043=1:2\\ \Rightarrow k=1;t=2\\ \Rightarrow CTHH.X:NO_2\\ Cu+4HNO_3\rightarrow Cu\left(NO_3\right)_2+2NO_2+2H_2O\\ n_{HNO_3}=\dfrac{4}{2}.n_{NO_2}=\dfrac{4}{2}.0,03=0,06\left(mol\right)\\ \Rightarrow m_{ddHNO_3}=\dfrac{0,06.63.100}{40}=9,45\left(g\right)\)

bn tk luôn tính độ mol hén

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{HCl}=2.n_{Zn}=2.0,1=0,2\left(mol\right)\\a, \Rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\\ b,n_{H_2}=_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow m_{ZnCl_2}=136.0,1=13,6\left(g\right)\\ c,Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ n_{Fe_2O_3}=\dfrac{n_{H_2}}{3}=\dfrac{0,1}{3}=\dfrac{1}{30}\left(mol\right)\\ \Rightarrow m_{Fe_2O_3}=\dfrac{1}{30}.160=\dfrac{16}{3}\left(g\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

a: \(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)

\(\Leftrightarrow n_{ZnCl_2}=0.1\left(mol\right)\)

\(\Leftrightarrow n_{HCl}=0.2\left(mol\right)\)

\(m_{HCl}=0.2\cdot36.5=7.3\left(g\right)\)

b: \(m_{ZnCl_2}=0.1\cdot136=13.6\left(g\right)\)