Cách 2:

Theo gt: p + e + n = 40

mà p = e

\(\Rightarrow2p+n=40\)

\(\Rightarrow n=40-2p\) (1)

mà \(p\le n\le1,5p\left(2\right)\)

(1)(2) \(\Rightarrow p\le40-2p\le1,5p\)

\(\Rightarrow\left\{{}\begin{matrix}40-2p\ge p\\40-2p\le1,5p\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}p\le13,33\\p\ge11,42\end{matrix}\right.\)

\(\Rightarrow11,42\le p\le13,33\)

loại loại nhận

Vậy Y là Nhôm ( Al )

Có a; b; c tỉ lậ với 2; 3; 4

=> \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)

=> \(\frac{a}{2}=\frac{2b}{6}=\frac{3c}{12}=\frac{a+2b+3c}{2+6+12}=\frac{-20}{20}=-1\)

=> a = (-1).2 = -2

     2b = (-1).6 = (-6) => b = -3

     3c = (-1).12 = (-12) => c = -4

KL: a = -2; b = -3; c = -4

6. \(n_{NaOH}=\dfrac{200.10\%}{40}=0,5\left(mol\right)\)

\(n_{HCl}=2.0,2=0,4\left(mol\right)\)

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

0,5mol 0,4mol

Lập tỉ số: \(n_{NaOH}\) : \(n_{HCl}=0,5>0,4\)

\(\Rightarrow NaOH\) dư; HCl hết

\(n_{NaOH\left(dư\right)}=0,5-0,4=0,1\left(mol\right)\)

\(m_{NaOH\left(dư\right)}=0,1.40=4\left(g\right)\)

7. \(n_{Ba\left(OH\right)_2}=\dfrac{300.30\%}{171}=0,52\left(mol\right)\)

Pt: \(Ba\left(OH\right)_2+FeSO_4\rightarrow BaSO_4+Fe\left(OH\right)_2\)

0,52mol \(\rightarrow0,52mol\)\(\rightarrow0,52mol\)

\(m_{BaSO_4}=0,52.233=121,16\left(g\right)\)

\(m_{Fe\left(OH\right)_2}=0,52.90=46,8\left(g\right)\)

\(\Sigma_{hh\left(spu\right)}=300+800-121,16=978,84\left(g\right)\)

\(C\%_{Fe\left(OH\right)_2}=\dfrac{46,8.100}{978,84}=4,78\%\)

5. \(n_{BaCl_2}=\dfrac{400.5,2\%}{208}=0,1\left(mol\right)\)

\(m_{dd}=1,14.100=114\left(g\right)\)

\(m_{H_2SO_4}=114.20\%=22,8\left(g\right)\)

\(n_{H_2SO_4}=\dfrac{22,8}{98}=0,23\left(mol\right)\)

\(Pt:BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)

0,1 mol 0,23mol \(\rightarrow0,1mol\) \(\rightarrow0,2mol\)

Lập tỉ số: \(n_{BaCl_2}\) : \(n_{H_2SO_4}=0,1< 0,23\)

\(\Rightarrow BaCl_2\) hết; \(H_2SO_4\) dư

\(m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

\(n_{H_2SO_4\left(dư\right)}=0,23-0,1=0,13\left(mol\right)\)

\(\Sigma_{hh\left(spu\right)}=400+114-23,3=490,7\left(g\right)\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,13.98.100}{490,7}=2,6\%\)

\(C\%_{HCl}=\dfrac{0,2.36,5.100}{490,7}=1,49\%\)

4.\(n_{HCl}=\dfrac{7,3.20\%}{36,5}=0,04\left(mol\right)\)

\(n_{AgNO_3}=\dfrac{50.15\%}{170}=0,044\left(mol\right)\)

Pt: \(AgNO_3+HCl\rightarrow AgCl+HNO_3\)

0,044mol 0,04mol \(\rightarrow\) 0,04mol

Lập tỉ số: \(n_{AgNO_3}\) : \(n_{HCl}=0,044>0,04\)

\(\Rightarrow AgNO_3\) dư; HCl hết

\(m_{AgCl}=0,04.143,5=5,74\left(g\right)\)

3. \(n_{H_2SO_4}=2.0,4=0,8\left(mol\right)\)

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

1,6mol \(\leftarrow\) 0,8mol

\(m_{NaOH}=1,6.40=64\left(g\right)\)

\(m_{dd}=\dfrac{64.100}{20}=320\left(g\right)\)

b) \(Pt:Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)

0,8mol \(\leftarrow\) 0,8mol

\(m_{Ca\left(OH\right)_2}=0,8.74=59,2\left(g\right)\)

\(m_{dd}=\dfrac{59,2.100}{7,4}=800\left(g\right)\)

\(V_{Ca\left(OH\right)_2}=\dfrac{800}{1,05}=761,9\left(ml\right)\)

1. \(n_{ZnO}=\dfrac{2,835}{81}=0,035\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{140.20\%}{98}=0,285\left(mol\right)\)

\(Pt:ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

0,035mol 0,285mol \(\rightarrow\) 0,035 mol

Lập tỉ số: n_ZnO : \(n_{H_2SO_4}=0,035< 0,285\)

\(\Rightarrow ZnO\) hết; \(H_2SO_4\) dư

\(n_{H_2SO_4\left(dư\right)}=0,0285-0,035=0,25\left(mol\right)\)

\(m_{H_2SO_4\left(dư\right)}=0,25.98=24,5\left(g\right)\)

\(m_{ZnSO_4}=0,035.161=5,635\left(g\right)\)

\(\Sigma_{hh\left(spu\right)}=2,835+140=142,835\left(g\right)\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{24,5.100}{142,835}=17,15\%\)

\(C\%_{ZnSO_4}=\dfrac{5,635.100}{142,835}=4\%\)

PTK của 4 nguyên tử H : 4.16 = 64

PTK = X + 4.2 = X +8

\(\Rightarrow X+8=64\)

\(\Rightarrow X=56\)

X là Sắt ( Fe )

b) \(\%Fe=\dfrac{56.100}{64}=87,5\%\)

\(n_{H_2}=\dfrac{0,784}{22,4}=0,035\left(mol\right)\)

\(Pt:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (1)

\(Pt:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\) (2)

\(Pt:Fe+H_2SO_4\rightarrow FeSO_4+H_2\) (3)

(1)(2)(3) \(\Rightarrow n_{H_2SO_4}=n_{H_2}=0,035\left(mol\right)\)

m_muối = \(m_{KL}+m_{axit}-m_{H_2}\)

= 2,48 + 0,035.98 - 0,035.2 = 5,84 (g)

Ít ra tui đã đi học được 1-2 tuần rồi =))