Bài 1. Tính chất hóa học của oxit. Khái quát về sự phân loại oxit

$n_{Fe_2O_3} = \dfrac{16}{160} = 0,1(mol)$
$n_{H_2SO_4} = \dfrac{150.2,94\%}{98} = 0,045(mol)$
$Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O$
Ta thấy : 

$n_{Fe_2O_3} : 1 > n_{H_2SO_4} : 3$ nên $Fe_2O_3$ dư

$n_{Fe_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = 0,015(mol)$
$m_{Fe_2(SO_4)_3} = 0,015.400 = 6(gam)$

đề hỏi gì vậy bn :) ?

\(n_{NaOH}=0,2.2=0,4\left(mol\right)\); \(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

\(n_{HCl}=\dfrac{175.14,6\%}{36,5}=0,7\left(mol\right)\)

PTHH: \(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\)

                0,3<-----0,3<--------------0,3<----0,45

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

                 0,1---->0,1

             \(NaAlO_2+HCl+H_2O\rightarrow NaCl+Al\left(OH\right)_3\downarrow\)

                   0,3----->0,3---------------------->0,3

             \(Al\left(OH\right)_3+3HCl\rightarrow AlCl_3+3H_2O\)

                   0,1<------0,3

=> \(n_{Al\left(OH\right)_3}=0,3-0,1=0,2\left(mol\right)\Rightarrow m=0,2.78=15,6\left(g\right)\)

a)

PTHH: \(Fe_xO_y+yH_2\underrightarrow{t^o}xFe+yH_2O\)

\(n_{H_2O}=\dfrac{7,2}{18}=0,4\left(mol\right)\) => n_{O(mất đi)} = 0,4 (mol)

\(n_{Fe\left(A\right)}=\dfrac{28,4.59,155\%}{56}=0,3\left(mol\right)\)

Xét n_Fe(A) : n_{O(mất đi)} = 0,3 : 0,4 = 3 : 4

=> CTHH: Fe₃O₄

b)

Do sau pư thu được muối sunfat => A là muối sunfat

B là M(NO₃)_m

\(n_{M\left(NO_3\right)_m}=\dfrac{50.10,44\%}{M_M+62m}\left(mol\right)\)

Bảo toàn M: 

 \(n_{M\left(NO_3\right)_m}=2.n_{M_2\left(SO_4\right)_m}\)

=> \(\dfrac{5,22}{M_M+62m}\) = \(2.\dfrac{4,66}{2.M_M+96m}\)

=> 10,44.M_M + 501,12.m = 9,32.M_M + 577,84m

=> M_M = 68,5m (g/mol)

Xét m = 2 thỏa mãn => M_M = 137 (g/mol)

=> M là Ba

B là Ba(NO₃)₂

\(n_{Ba\left(NO_3\right)_2}=\) 0,02 (mol)

m_A = 200.1,36% = 2,72 (g)

TH1: A là X(HSO₄)_n

PTHH: \(X\left(HSO_4\right)_n+nBa\left(NO_3\right)_2\rightarrow nBaSO_4+X\left(NO_3\right)_n+nHNO_3\)

                \(\dfrac{0,02}{n}\)<--------0,02

=> \(M_{X\left(HSO_4\right)_n}=\dfrac{2,72}{\dfrac{0,02}{n}}=136n\left(g/mol\right)\Rightarrow M_X=39n\left(g/mol\right)\)

Xét n = 1 thỏa mãn => M_X = 39 (g/mol) => X là K

A là KHSO₄

TH2: B là X₂(SO₄)_n

PTHH: \(X_2\left(SO_4\right)_n+nBa\left(NO_3\right)_2\rightarrow nBaSO_4+2X\left(NO_3\right)_n\)

               \(\dfrac{0,02}{n}\)<----------0,02

=> \(M_{X_2\left(SO_4\right)_n}=\dfrac{2,72}{\dfrac{0,02}{n}}=136n\left(g/mol\right)\Rightarrow M_X=20n\left(g/mol\right)\)

Xét n = 2 thỏa mãn => M_X = 40 (g/mol)

=> X là Ca (Loại do CaSO₄ kết tủa)

ý B là khối lg oxi hay oxit em ?

a)K2O+H2O->2KOH

=>m K2O=4g=>n K2O=\(\dfrac{2}{47}\)mol

=>n MgO=0,05 mol

=>%O =\(\dfrac{\dfrac{2}{47}.16+0,05.16}{6}.100=24,68\%\)

$m_{dd\ sau\ pư} = 9,18 + 98,55 = 107,73(gam)$
$\Rightarrow m_{muối} = 107,73.22,306\% = 24,03(gam)$

$R_2O_3 + 6HCl \to 2RCl_3 + 3H_2O$
Theo PTHH :

$2n_{R_2O_3} = n_{RCl_3}$
$\Rightarrow \dfrac{2.9,18}{2R + 16.3} = \dfrac{24,03}{R + 35,5.3}$

$\Rightarrow R = 27(Al)$
$C\% = \dfrac{0,09.6.36,5}{98,55}.100\% = 20\%$

Tên : Nhôm oxit

\(a.Ba\left(OH\right)_2+K_2CO_3\rightarrow BaCO_3+2KOH\\ n_{Ba\left(OH\right)_2}=\dfrac{50.17,1}{171}=0,05\left(mol\right)\\ n_{K_2CO_3}=n_{BaCO_3}=n_{Ba\left(OH\right)_2}=0,05\left(mol\right)\\ n_{KOH}=2n_{Ba\left(OH\right)_2}=0,1\left(mol\right)\\ m_{ddK_2CO_3}=\dfrac{0,05.138}{13,8\%}=50\left(g\right)\\ m_{ddsaupu}=50+50-0,05.197=90,15\left(g\right)\\ C\%_{KOH}=\dfrac{0,1.56}{90,15}.100=6,21\%\)

\(a.2KOH+CuSO_4\rightarrow K_2SO_4+Cu\left(OH\right)_2\\ b.n_{KOH}=\dfrac{70.11,2\%}{56}=0,14\left(mol\right) \\ n_{CuSO_4}=n_{Cu\left(OH\right)_2}=n_{K_2SO_4}=\dfrac{1}{2}n_{KOH}=0,07\left(mol\right)\\ V_{ddCuSO_4}=\dfrac{0,07}{1}=0,07\left(l\right)=70ml \\ m_{ddCuSO_4}=70.1,31=91,7\left(g\right)\\ m_{ddsaupu}=70+91,7-0,07.98=154,84\left(g\right)\\ C\%_{K_2SO_4}=\dfrac{0,07.174}{154,84}.100=7,87\%\)

a)b) 

\(m_{ct\left(KOH\right)}=\dfrac{m_{dd}\cdot C\%}{100\%}=\dfrac{70\cdot11,2\%}{100\%}=7,84\left(g\right)\)

\(n_{ct\left(KOH\right)}=\dfrac{7,84}{39+16+1}=0,14\left(mol\right)\)

PTHH:

\(2KOH+CuSO_4->Cu\left(OH\right)_2+K_2SO_4\)

0,14    --> 0,07     -->        0,07 -->   0,07    (mol)

\(V_{CuSO_4}=\dfrac{n}{C_M}=\dfrac{0,07}{1}=0,07\left(lít\right)=70\left(ml\right)\)

\(m_{dd\left(CuSO_4\right)}=d.V=1,31\cdot70=91,7\left(g\right)\)

\(m\left(ddsau\right)=m_{KOH}+m_{CuSO_4}-m_{Cu\left(OH\right)_2}\\ =70+91,7-6,86=154,84\left(g\right)\)

Chất thu được sau phản ứng là \(K_2SO_4\)

\(C\%_{K_2SO_4}=\dfrac{m_{ct}}{mddsau}\cdot100\%=\dfrac{0,07\cdot174}{154,84}\cdot100\%\approx7,87\%\)

a)

$m_{dd} = 0,8.342 + 526,4 = 800(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,8.342}{800}.100\% = 34,2\%$

b)

$V_{dd} =\dfrac{m_{dd}}{D} = \dfrac{800}{1,25} = 640(ml) = 0,64(lít)$

$C_{M_{Al_2(SO_4)_3}} = \dfrac{0,8}{0,64} = 1,25M$