a) 

Số n = \(\dfrac{95-25}{2}=35\left(hạt\right)\)

Số p = số e = \(\dfrac{95-35}{2}=30\left(hạt\right)\)

R là Zn 

Zn chứa 30 hạt proton, 30 hạt electron, 35 hạt nơtron

b)

\(NTK_X=\dfrac{4.65}{6,5}=40\left(đvC\right)\)

=> X là Ca

Khối lượng nguyên tử Ca là: 40.0,16605.10^-23 = 6,642.10^-23 (g)

đề bảo là "500ml dung dịch XOH chứa 20g NaOH" mà bn :) ?

1 lít đâu ra vậy :) ?

a)

Gọi \(\left\{{}\begin{matrix}n_{HCl}=3a\left(mol\right)\\n_{H_2SO_4}=a\left(mol\right)\end{matrix}\right.\)

\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

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

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

Theo PTHH: n_NaOH = 3a + 2a = 0,5 (mol)

=> a = 0,1 

\(\left\{{}\begin{matrix}C_{M\left(HCl\right)}=\dfrac{0,3}{0,1}=3M\\C_{M\left(H_2SO_4\right)}=\dfrac{0,1}{0,1}=1M\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}n_{HCl}=3.0,2=0,6\left(mol\right)\\n_{H_2SO_4}=1.0,2=0,2\left(mol\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}n_{NaOH}=0,001.V.0,2=0,0002V\left(mol\right)\\n_{Ba\left(OH\right)_2}=0,001.V.0,1=0,0001V\left(mol\right)\end{matrix}\right.\)

Gọi công thức chung của axit là HA

n_HA = 0,6 + 0,2.2 = 1 (mol)

Gọi công thức chung của bazo là BOH

n_BOH = 0,0002V + 2.0,0001V = 0,0004V (mol)

PTHH: \(BOH+HA\rightarrow BA+H_2O\)

Theo PTHH: n_BOH = 0,0004V = 1

=> V = 2500 (ml)

\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)

\(Ba\left(OH\right)_2+H_2CO_3\rightarrow BaCO_3+2H_2O\)

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

PTHH: \(2yAl+3Fe_xO_y\rightarrow yAl_2O_3+3xFe\)

=> \(\dfrac{n_{Al_2O_3}}{n_{Fe}}=\dfrac{y}{3x}\)

- Phần 1: Gọi \(\left(n_{Al};n_{Fe};n_{Al_2O_3}\right)=\left(a;b;c\right)\)

=> 27a + 56b + 102c = 7,245 (1)

PTHH: \(2Al+6H_2SO_{4\left(đ,n\right)}\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)

              a------------------------------------>1,5a

            \(2Fe+6H_2SO_{4\left(đ,n\right)}\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

               b--------------------------------------->1,5b

=> 1,5a + 1,5b = 0,12375

=> a + b = 0,0825 (2)

- Phần 2: Gọi \(\left(n_{Al};n_{Fe};n_{Al_2O_3}\right)=\left(ak;bk;ck\right)\)

Chất rắn không tan là Fe

\(bk=\dfrac{1,26}{56}=0,0225\left(mol\right)\)

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

          0,005<------------------------------------0,0075

=> ak = 0,005

=> a = \(\dfrac{2}{9}\)b (3)

(1)(2)(3) => a = 0,015 (mol); b = 0,0675 (mol); c = 0,03 (mol); k = \(\dfrac{1}{3}\)

Ta có: \(\dfrac{y}{3x}=\dfrac{0,03}{0,0675}\Rightarrow\dfrac{x}{y}=\dfrac{3}{4}\) => Fe₃O₄

\(m=\left(1+\dfrac{1}{3}\right)\left(0,015.27+0,0675.56+0,03.102\right)=\) 9,66 (g)

1) 

\(n_{Na_2CO_3}=\dfrac{100.16,96\%}{106}=0,16\left(mol\right)\)

\(n_{BaCl_2}=\dfrac{200.10,4\%}{208}=0,1\left(mol\right)\)

PTHH: \(BaCl_2+Na_2CO_3\rightarrow BaCO_3+2NaCl\)

Xét tỉ lệ: \(\dfrac{0,16}{1}>\dfrac{0,1}{1}\) => Na₂CO₃ dư, BaCl₂ hết

PTHH: \(BaCl_2+Na_2CO_3\rightarrow BaCO_3+2NaCl\)

                0,1------->0,1--------->0,1------>0,2

m_{dd sau pư} = 100 + 200 - 0,1.197 = 280,3 (g)

\(\left\{{}\begin{matrix}C\%_{Na_2CO_3}=\dfrac{\left(0,16-0,1\right).106}{280,3}.100\%=2,27\%\\C\%_{NaCl}=\dfrac{0,2.58,5}{280,3}.100\%=4,174\%\end{matrix}\right.\)

2)

CTHH của oxit là M₂O_n

Giả sử hòa tan 100 (g) oxit

\(n_{M_2O_n}=\dfrac{100}{2.M_M+16n}\left(mol\right)\)

PTHH: \(M_2O_n+2nHCl\rightarrow2MCl_n+nH_2O\)

            \(M_2O_n+2HNO_3\rightarrow2M\left(NO_3\right)_n+nH_2O\)

Theo PTHH: \(n_{MCl_n}=n_{M\left(NO_3\right)_n}=2.n_{M_2O_n}=\dfrac{100}{M_M+8n}\left(mol\right)\)

Ta có: \(m_{M\left(NO_3\right)_n}-m_{MCl_n}=\dfrac{100}{M_M+8n}\left(M_M+62n-M_M-35,5n\right)\)

=> \(\dfrac{100}{M_M+8n}.26,5n=100.99,38\%=99,38\)

=> \(M_M\approx18,665n\left(g/mol\right)\)

Xét n = 3 thỏa mãn => M_M = 56 (g/mol)

=> M là Fe

CTHH: Fe₂O₃

\(n_{CuSO_4}=0,3.2=0,6\left(mol\right)\)

Gọi số mol CuSO₄ pư là a (mol)

PTHH: \(Fe+CuSO_4\rightarrow FeSO_4+Cu\)

             a<--------a---------->a----->a

m_tăng = 64a - 56a = 2,4

=> a = 0,3 (mol)

\(\left\{{}\begin{matrix}C_{M\left(CuSO_4\right)}=\dfrac{0,6-0,3}{0,3}=1M\\C_{M\left(FeSO_4\right)}=\dfrac{0,3}{0,3}=1M\end{matrix}\right.\)

Gọi số mol CuSO₄ là x (mol)

PTHH: \(Fe+CuSO_4\rightarrow FeSO_4+Cu\)

             x<--------x------------------>x

Ta có: 50 - 56x + 64x = 51

=> x = 0,125 (mol)

=> \(C_{M\left(CuSO_4\right)}=\dfrac{0,125}{0,1}=1,25M\)

a)

\(n_{NaOH}=\dfrac{200.6\%}{40}=0,3\left(mol\right)\)

\(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

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

               0,04<----0,04<----------------0,04---0,06

            \(2NaOH+Al_2O_3\rightarrow2NaAlO_2+H_2O\)

                0,26---->0,13------>0,26

             \(3Fe_3O_4+8Al\underrightarrow{t^o}4Al_2O_3+9Fe\)

            0,0975<---0,26<--0,13--->0,2925

\(\left\{{}\begin{matrix}m_{Fe_3O_4\left(X\right)}=0,0975.232=22,62\left(g\right)\\m_{Al\left(X\right)}=\left(0,04+0,26\right).27=8,1\left(g\right)\end{matrix}\right.\)

b) m_Z = 200 + 0,04.27 + 0,13.102 - 0,06.2 = 214,22 (g)

\(C\%_{NaAlO_2}=\dfrac{\left(0,04+0,26\right).82}{214,22}.100\%=11,5\%\)

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

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

             \(Fe+2HCl\rightarrow FeCl_2+H_2\)

         0,2925----------------->0,2925

=> V = (0,06 + 0,2925).22,4 = 7,896 (l)

1)

\(n_{SO_2}=\dfrac{2,688}{22,4}=0,12\left(mol\right)\)

PTHH: \(2Fe+6H_2SO_{4\left(đ,n\right)}\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

           0,08<-----------------------------------0,12

=> n_Fe = 0,08 (mol)

\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

\(n_{Al_2O_3}=\dfrac{5,1}{102}=0,05\left(mol\right)\)

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

                 0,02<-- 0,02<----------------0,02<---0,03

            \(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)

              0,04<----0,08--------0,08

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

                 0,1<-----------------------------------0,1

            \(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)

                 0,1<------0,05

=> \(n_{Al_2O_3}=0,04\left(mol\right)\Rightarrow n_O=0,12\left(mol\right)\)

Xét n_Fe : n_O = 0,08 : 0,12 = 2 : 3

=> Fe₂O₃

m = 0,02.27 + 0,04.102 + 0,08.56 = 9,1 (g)

2)

C gồm \(\left\{{}\begin{matrix}NaAlO_2:0,1\left(mol\right)\\NaOH:x\left(mol\right)\end{matrix}\right.\)

n_HCl = 0,2.1 = 0,2 (mol)

\(n_{Al\left(OH\right)_3}=\dfrac{6,24}{78}=0,08\left(mol\right)\)

TH1: Kết tủa không bị hòa tan

PTHH: NaOH + HCl --> NaCl + H₂O

                x------>x

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

                                            0,08<------------0,08

=> x + 0,08 = 0,2

=> x = 0,12 (mol)

n_NaOH(bđ) = 0,02 + 0,08 + 0,12 = 0,22 (mol)

=> m_NaOH(bđ) = 0,22.40 = 8,8 (g)

TH2: Kết tủa bị hòa tan 2 phần

PTHH: NaOH + HCl --> NaCl + H₂O

                x------>x

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

                  0,1---------------->0,1-------------->0,1

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

                   0,02----->0,06

=> x + 0,1 + 0,06 = 0,2

=> x = 0,04 (mol)

=> n_{NaOH (bđ)} = 0,04 + 0,02 + 0,08 = 0,14 (mol)

=> m_NaOH(bđ) = 0,14.40 = 5,6 (g)