a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

            0,2--->0,6------------>0,3

=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

b) \(C_{M\left(HCl\right)}=\dfrac{0,6}{0,25}=2,4M\)

a)

Do khối lượng chất tan bằng nhau, lượng nước thêm vào bằng nhau

=> Khối lượng dd bằng nhau

=> C% 2 dd bằng nhau

b)

\(n_{CuSO_4}=\dfrac{a}{160}\left(mol\right)\); \(n_{KNO_3}=\dfrac{a}{101}\left(mol\right)\)

Do \(n_{CuSO_4}< n_{KNO_3}\), thể tích 2 dd bằng nhau

=> \(C_{M\left(CuSO_4\right)}< C_{M\left(KNO_3\right)}\)

c)

Nếu \(n_{CuSO_4}=n_{KNO_3}\left(=a\left(mol\right)\right)\), thể tích 2 dd bằng nhau

=> \(C_{M\left(CuSO_4\right)}=C_{M\left(KNO_3\right)}\)

300cm³ đâu ra vậy ?

Có: \(\left\{{}\begin{matrix}2p+n=46\\p\le n\le1,5p\end{matrix}\right.\)

=> \(\dfrac{92}{7}\le p\le\dfrac{46}{3}\)

=> \(\left[{}\begin{matrix}p=14\\p=15\end{matrix}\right.\)

- Nếu p = 14 => Điện tích hạt nhân = 14+

- Nếu p = 15 => Điện tích hạt nhân = 15+

Giả sử có 100a gam hợp chất

\(\rightarrow\left\{{}\begin{matrix}m_X=60\%.100a=60a\left(g\right)\\m_Y=100a-60a=40a\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}M_X=\dfrac{60a}{1}=60a\left(g\text{/}mol\right)\\M_Y=\dfrac{40a}{2}=20a\left(g\text{/}mol\right)\end{matrix}\right.\)

CTHH B: \(X_xY_y\left(x,y\in N\text{*}\right)\)

\(\%Y=100\%-75\%=25\%\)

\(\rightarrow x:y=\dfrac{75}{60a}:\dfrac{25}{20a}=1:1\)

Vì \(x,y\in N\text{*}\rightarrow x=y=1\)

Vậy CTHH của B là \(XY\)

- Xét hợp chất A 

Có: \(\%X=\dfrac{NTK_X}{NTK_X+2.NTK_Y}.100\%=60\%\)

=> 0,4.NTK_X = 1,2.NTK_Y

=> NTK_X = 3.NTK_Y

- Xét hợp chất B:

CTPT: X_uY_v

Có: \(\%X=\dfrac{u.NTK_X}{u.NTK_X+v.NTK_Y}.100\%=75\%\)

=> \(\dfrac{u.3.NTK_Y}{u.3.NTK_Y+v.NTK_Y}=0,75\)

=> \(\dfrac{3u}{3u+v}=0,75\)

=> 0,75u = 0,75v

=> u = v => u : v = 1 : 1

=> B có CTPT là XY

\(n_{NH_3}=\dfrac{22,4}{22,4}=1\left(mol\right)\\ \rightarrow m_{NH_3}=1.17=17\left(g\right)\\ \rightarrow C\%_{NH_3}=\dfrac{17}{200}.100\%=8,5\%\)

Câu 1:

a)

\(M_{XH_4}=\dfrac{4.100}{25}=16\left(g\text{/}mol\right)\\ \rightarrow M_X=\dfrac{16-4}{1}=12\left(g\text{/}mol\right)\)

=> X là C

\(M_{YO_2}=\dfrac{16.2.100}{50}=64\left(g\text{/}mol\right)\\ M_Y=\dfrac{64-16.2}{1}=32\left(g\text{/}mol\right)\)

=> Y là S

b) 

\(d_{SO_2\text{/}CH_4}=\dfrac{64}{16}=4\left(lần\right)\)

c)

Đặt \(m_{SO_2}=m_{CH_4}=a\left(g\right)\)

\(\rightarrow\left\{{}\begin{matrix}n_{CH_4}=\dfrac{a}{16}\left(mol\right)\\n_{SO_2}=\dfrac{a}{64}\left(mol\right)\end{matrix}\right.\\ \rightarrow\dfrac{V_1}{V_2}=\dfrac{\dfrac{a}{16}}{\dfrac{a}{64}}=\dfrac{4}{1}\)

Hay \(V_1:V_2=4:1\)

Câu 2:

Chất rắn không tan trong dd HCl là Cu

\(\rightarrow m_{Cu}=1,86\left(g\right)\)

\(n_{H_2}=\dfrac{3,024}{22,4}=0,135\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)

\(\rightarrow56a+27b+1,86=6\left(1\right)\)

PTHH: 

\(Fe+2HCl\rightarrow FeCl_2+H_2\left(1\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\left(2\right)\)

Theo PT \(\left(1\right),\left(2\right):n_{H_2}=n_{Fe}+\dfrac{3}{2}n_{Al}\left(mol\right)\)

\(\rightarrow a+\dfrac{3}{2}b=0,135\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\rightarrow\left\{{}\begin{matrix}a=0,045\left(mol\right)\\b=0,06\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{1,86}{6}.100\%=31\%\\\%m_{Fe}=\dfrac{0,045.56}{6}.100\%=42\%\\\%m_{Al}=100\%-31\%-42\%=27\%\end{matrix}\right.\)

Câu 3:

a) 

Gọi dd H₂SO₄ 60% là dd₁, dd H₂SO₄ 20% là dd₂

\(\left\{{}\begin{matrix}m_{dd1}=700.1,503=1052,1\left(g\right)\\m_{dd2}=500.1,143=571,5\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}m_{H_2SO_4\left(dd1\right)}=\dfrac{1052,1.60}{100}=631,26\left(g\right)\\m_{H_2SO_4\left(dd2\right)}=\dfrac{571,5.20}{100}=114,3\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\sum n_{H_2SO_4\left(ddA\right)}=\dfrac{631,26+114,3}{98}=7,6\left(mol\right)\\\sum V_{ddH_2SO_4\left(ddA\right)}=500+700=1200\left(ml\right)=1,2\left(l\right)\end{matrix}\right.\\ \rightarrow C_{M\left(H_2SO_4.A\right)}=\dfrac{7,61}{1,2}=\dfrac{19}{3}M\)

b)

\(n_{H_2}=\dfrac{2}{22,4}=\dfrac{5}{56}\left(mol\right)\\ \xrightarrow[]{\text{BTNT H}}n_{H_2SO_4}=\dfrac{5}{56}\left(mol\right)\\ \rightarrow C_{M\left(H_2SO_4.A\right)}=\dfrac{\dfrac{5}{56}}{0,2}=\dfrac{25}{56}M\\ \rightarrow V_{ddA\left(sau.khi.thêm.nước\right)}=\dfrac{7,6}{\dfrac{25}{56}}=17,024\left(l\right)\)

Câu 4:

a) 

\(n_{CO}=n_{CO_2}=n_{N_2}=n_{H_2}=\dfrac{204}{28+44+28+2}=2\left(mol\right)\\ \rightarrow m_{H_2}=2.2=4\left(g\right)\)

b)

\(n_{CO_2}=\dfrac{20,16}{22,4}=0,9\left(mol\right)\\ n_{Fe_2O_3}=\dfrac{80}{160}=0,5\left(mol\right)\)

PTHH: \(Fe_2O_3+3CO\xrightarrow[]{t^o}2Fe+3CO_2\)

            0,3<------0,9

\(\rightarrow H=\dfrac{0,3}{0,5}.100\%=60\%\)

Theo PT: \(n_O=n_{CO_2}=0,9\left(mol\right)\)

\(\xrightarrow[]{\text{BTNT}}m_{chất.rắn}=80-0,9.16=65,6\left(g\right)\)

$n_{Al} = \dfrac{9}{27} = \dfrac{1}{3} (mol)$

Đặt $n_{SO_2} = n_{H_2S} = a (mol)$

Quá trình nhường e: $Al^0 - 3e \rightarrow Al^{3+}$

                                    $\dfrac{1}{3}$------>1

Quá trình nhận e:

$S^{6+} + 2e \rightarrow S^{+4}$

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

$S^{+6} + 8e \rightarrow S^{2-}$

a--------->8a

$\xrightarrow{BTe} 2a + 8a = 1 $

$\Leftrightarrow a = 0,1 (mol)$

$\rightarrow V = (0,1 + 0,1).22,4 = 4,48 (l)$

$\rightarrow M_A = \dfrac{0,1.64 + 0,1.34}{0,2} = 49 (g/mol)$

$\rightarrow d_{A/kk} = \dfrac{49}{29} = 1,69$

Gọi CTHH của oxit là \(R_xO_y\left(x,y\in N\text{*},\text{2y/x là hoá trị của kim loại R}\right)\)

\(n_{HCl}=1,5.0,2=0,3\left(mol\right)\)

PTHH: \(R_xO_y+2yHCl\rightarrow xRCl_{2y\text{/}x}+yH_2O\)

              \(\dfrac{0,15}{y}\)<--0,3

\(\rightarrow n_R=xn_{R_xO_y}=x.\dfrac{0,15}{y}=\dfrac{0,15x}{y}\left(mol\right)\)

Theo PTHH: \(n_O=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,3=0,15\left(mol\right)\)

\(\xrightarrow[]{\text{BTNT}}m_R=8-0,15.16=5,6\left(g\right)\)

\(\rightarrow M_R=\dfrac{5,6}{\dfrac{0,15x}{y}}=\dfrac{112y}{3x}=\dfrac{56}{3}.\dfrac{2y}{x}\left(g\text{/}mol\right)\)

Vì 2y/x là hoá trị R nên ta có:

=> R là Fe

\(\rightarrow\dfrac{2y}{x}=3\Leftrightarrow\dfrac{x}{y}=\dfrac{2}{3}\)

Do \(x,y\in N\text{*}\rightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

Vậy CTHH của oxit là \(Fe_2O_3\)

\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\\ n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\\ pthh:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ LTL:\dfrac{0,1}{4}>\dfrac{0,1}{5}\) 
P dư 
\(n_{P\left(p\text{ư}\right)}=\dfrac{4}{5}n_{O_2}=0,08\left(mol\right)\\ m_{P\left(d\right)}=0,08.31=0,62\left(g\right)\\ n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=0,04\left(mol\right)\\ m_{P_2O_5}=0,04.142=5,68\left(g\right)\\ m_{sp}=0,62+5,68=6,3\left(g\right)\)