\(n_{Fe}=\dfrac{10,08}{56}=0,18\left(mol\right)\\ n_{NO}=\dfrac{3,024}{22,4}=0,135\left(mol\right)\\ Fe+4HNO_3\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\\ Fe_{dư}+2Fe\left(NO_3\right)_3\rightarrow3Fe\left(NO_3\right)_2\\ n_{Fe\left(dư\right)}=0,18-0,135=0,045\left(mol\right)\\ n_{Fe\left(NO_3\right)_2}=3.0,045=0,135\left(mol\right)\\ \Rightarrow m_{Fe\left(NO_3\right)_2}=180.0,135=24,3\left(g\right)\\ \Rightarrow Chọn.B\)

Coi n là hoá trị của kim loại M

$4M + nO_2 \xrightarrow{t^o} 2M_2O_n$

$M_2O_n + 2nHNO_3 \to 2M(NO_3)_n + nH_2O$

Bảo toàn khối lượng : $m_{O_2} = 9,6 - 7,68 = 1,92(gam)$
$\Rightarrow n_{O_2} = 0,06(mol)$

Theo PTHH : 

$n_{M_2O_n} = \dfrac{2}{n}n_{O_2} = \dfrac{0,12}{n}(mol)$
$n_{HNO_3} = 2n.n_{M_2O_n} = 0,24(mol)$
$C_{M_{HNO_3}} = \dfrac{0,24}{0,5} = 0,48M$
Đáp án C

x là hoá trị M (x:nguyên, dương)

\(4M+xO_2\rightarrow\left(t^o\right)2M_2O_x\\ n_{O_2}=\dfrac{9,6-7,68}{32}=0,06\left(mol\right)\Rightarrow n_M=\dfrac{0,24}{x}\left(mol\right)\\ \Rightarrow M_M=\dfrac{7,68}{\dfrac{0,24}{x}}=32x\left(\dfrac{g}{mol}\right)\\ Xét:x=1;x=2;x=3,..Chọn:x=2\left(TM\right)\\ \Rightarrow M_M=64\left(\dfrac{g}{mol}\right)\\ M:Đồng\left(Cu=64\right)\\ CuO+2HNO_3\rightarrow Cu\left(NO_3\right)_2+2H_2O\\ n_{HNO_3}=2.n_{oxit}=2.0,12=0,24\left(mol\right)\\ \Rightarrow C_{MddCu\left(NO_3\right)_2}=\dfrac{0,24}{0,5}=0,48\left(M\right)\\ \Rightarrow Chọn.C\)

$n_{N_2} = \dfrac{22,4}{22,4} = 1(mol)$
$n_{H_2} = \dfrac{24}{22,4} = \dfrac{15}{14}(mol)$
$N_2 + 3H_2 \buildrel{{t^o,p,xt}}\over\rightleftharpoons 2NH_3$

Ta thấy : 

$n_{N_2} : 1 > n_{H_2} : 3$ nên hiệu suất tính theo số mol của $H_2$

$n_{H_2\ pư} = \dfrac{15}{14}.30\% = \dfrac{9}{28}(mol)$
$n_{NH_3} = \dfrac{2}{3}n_{H_2\ pư} = \dfrac{3}{14}(mol)$
$V_{NH_3} = \dfrac{3}{14}.22,4 = 4,8(lít)$

Đáp án B

$n_O = \dfrac{86,3.19,47\%}{16} = 1,05(mol)$

Coi hỗn hợp gồm $Na(a\ mol) , Al(b\ mol)$ và O(1,05 mol)

Ta có : 

$23a + 27b + 1,05.16 = 86,3(1)$

Bảo toàn electron : $n_{Na} + 3n_{Al} = 2n_O + 2n_{H_2}$
$\Rightarrow a + 3b = 1,05.2 + 0,6.2(2)$

Từ (1)(2) suy ra $a = \dfrac{199}{70} ; b = \dfrac{16}{105}$

Suy ra Y gồm : 

$AlO_2^- : \dfrac{16}{105}(mol)$

$Na^+ : \dfrac{199}{70}(mol)$

$OH^- : \dfrac{199}{70} - \dfrac{16}{105} = \dfrac{113}{42}(mol)$

$n_{Al(OH)_3} = \dfrac{46,8}{78} = 0,6(mol)$

Suy ra : 

$n_{HCl} = n_{H^+} = n_{OH^-} + 4n_{AlO_2^-} - 3n_{Al(OH)_3} = 1,5(mol)$
$V_{dd\ HCl} = \dfrac{1,5}{2} = 0,75(lít)$

a) \(\left[Ca^{2+}\right]=0,015\left(M\right)\)

\(\left[OH^-\right]=0,015\cdot2=0,03\left(M\right)\)

b) \(\left[K^+\right]=0,2\left(M\right)\)

\(\left[MnO_4^-\right]=0,2\left(M\right)\)

\(K_{a_1}_{\left(H_3PO_4\right)}=10^{-2,148}\)

\(C^o_{H_3PO_4}=0,02M\)

A

Dung dịch sau điện phân có \(H^+\) nên hòa tan được \(\dfrac{3,2}{80}\) \(=0,04\) mol \(CuO\) 

\(\rightarrow n_{H+}=2n_{CuO}\)\(=0,08\left(mol\right)\)

\(n\)_{khí anot} \(=0,04\left(mol\right)\)

\(\left(-\right)\)Catot:

\(Cu^2+2e\rightarrow Cu\)

\(\left(+\right)\):Anot:

\(2Cl^-\rightarrow Cl_2+2e\)

\(2H_2O\rightarrow4H^++O2+4e\)

\(n_{O_2}=\dfrac{n_{H^+}}{4}=0,02\left(mol\right)\)

\(\rightarrow n_{Cl_2}=0,04-0,02=0,02\left(mol\right)\)

\(\rightarrow n_{NaCl}=0,02.2=0,04\left(mol\right)\)

BTe:\(2n_{Cu}=2n_{Cl_2}+4n_{O_2}\)

\(\rightarrow n_{CuSO_4}=n_{Cu}=0,06\left(mol\right)\)

Vậy \(m=0,06.160+0,04.58,5=11,94g\)

thanks

a) $n_{HCl} = 0,1.2 = 0,2(mol)$

$Fe_xO_y + 2yHCl \to xFeCl_{\dfrac{2y}{x}} + yH_2O$

Theo PTHH : 

$n_{Fe_xO_y} = \dfrac{1}{2y}n_{HCl} = \dfrac{0,1}{y}(mol)$
$\Rightarrow \dfrac{0,1}{y}.(56x + 16y) = 5,8$

$\Rightarrow \dfrac{x}{y} = \dfrac{3}{4}$

Vậy CTHH của oxit là $Fe_3O_4$
$Fe_3O_4 + 4CO \xrightarrow{t^o} 3Fe + 4CO_2$
$Fe_3O_4 + 4C \xrightarrow{t^o} 3Fe + 4CO$
$Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O$

Bài 1 :

$PTK_{NH_4NO_3} = 80(đvC)$ 

$\%N  = \dfrac{14.2}{80}.100\% = 35\%$

$\%H = \dfrac{1.4}{80}.100\% = 5\%$

$\%O = \dfrac{16.3}{80}.100\% = 60\%$

Bài 2 : 

a) $n_{Fe} = \dfrac{2,8}{56} = 0,05(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{HCl} =2 n_{Fe} = 0,1(mol)$
$V_{dd\ HCl} = \dfrac{0,1}{2} = 0,05(lít)$

b) $n_{H_2} = n_{Fe} = 0,05(mol)$
$V_{H_2} = 0,05.22,4 = 1,12(lít)$

thank bro 

a) PTHH:
`AO + H_2SO_4 -> ASO_4 + H_2O`

`B_2O_3 + 3H_2SO_4 -> B_2(SO_4)_3 + 3H_2O`

Theo pp tăng giảm khối lượng:

\(n_{SO_4}=n_O=\dfrac{50,2-18,2}{96-16}=0,4\left(mol\right)\)

`=>` \(n_{H_2SO_4}=n_{SO_4}=0,4\left(mol\right)\)

`=>` \(C\%_{H_2SO_4}=\dfrac{0,4.98}{400}.100\%=9,8\%\)

b) Ta có: \(n_{CuSO_4}=0,2.0,3=0,06\left(mol\right)\)

PTHH: \(3CuSO_4+2B\rightarrow B_2\left(SO_4\right)_3+3Cu\)

             0,06------>0,04------------------->0,06

`=> 0,06.64 - 0,04M_B = a + 2,76 - a`

`<=> M_B = 27`

`=>` B là Al

Gọi \(\left\{{}\begin{matrix}n_{AO}=2x\left(mol\right)\\n_{Al_2O_3}=x\left(mol\right)\end{matrix}\right.\) 

`=> x + 3x = 0,4`

`=> x = 0,1`

`=> 0,1.2.(M_A + 16) + 0,1.102 = 18,2`

`=> M_A = 24`

`=>` A là Mg

a)

$AO + H_2SO_4 \to ASO_4 + H_2O$
$B_2O_3 + 3H_2SO_4 \to B_2(SO_4)_3 + 3H_2O$

Gọi $n_{H_2O} = n_{H_2SO_4} = x(mol)$
Bảo toàn khối lượng : $18,2 + 98x = 50,2 + 18x$
$\Rightarrow x = 0,4(mol)$
$C\%_{H_2SO_4} = \dfrac{0,4.98}{400}.100\% = 9,8\%$

b) $2B + 3CuSO_4 \to B_2(SO_4)_3 + 3Cu$

Theo PTHH : 

$n_{Cu} = n_{CuSO_4} = 0,2.0,3 = 0,06(mol)$

Theo PTHH : $n_B = \dfrac{2}{3}n_{CuSO_4} = 0,04(mol)$

$\Rightarrow m_{tăng} = 0,06.64 - 0,04.B = a + 2,76 - a$

$\Rightarrow B = 27(Al)$

Gọi $n_{AO} = 2y(mol) ; n_{Al_2O_3} = y(mol)$

Theo PTHH : $n_{H_2SO_4} = n_{AO} + 3n_{Al_2O_3} = 5y = 0,4(mol)$
$\Rightarrow y = 0,08$

Suy ra : $0,08.2.(A + 16) + 0,08.102 = 18,2$

$\Rightarrow A = 46,75$

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