a)

PT ion đầy đủ: \(H^++NO_3^-+Li^++OH^-\rightarrow Li^++NO_3^-+H_2O\)

PT ion rút gọn: \(H^++OH^-\rightarrow H_2O\)

b)

PT ion đầy đủ: \(NH_4^++Cl^-+Cs^++OH^-\rightarrow Cs^++Cl^-+NH_3+H_2O\)

PT ion rút gọn: \(NH_4^++OH^-\rightarrow NH_3+H_2O\)

c)

PT ion đầy đủ: \(3Cu+8H^++8NO_3^-\rightarrow3Cu^{2+}+6NO_3^-+2NO+4H_2O\)

PT ion rút gọn: \(3Cu+8H^++2NO_3^-\rightarrow3Cu^{2+}+2NO+4H_2O\)

d)

PT ion đầy đủ: \(3H^++3Cl^-+3Na^++BO_3^{3-}\rightarrow3Na^++3Cl^-+H_3BO_3\)

PT ion rút gọn: \(3H^++BO_3^{3-}\rightarrow H_3BO_3\)

a) \(HNO_3+LiOH\rightarrow LiNO_3+H_2O\)

- Pt ion đầy đủ: \(H^++NO_3^-+Li^++OH^-\rightarrow Li^++NO_3^-+H_2O\)

- Pt ion rút gọn: \(H^++OH^-\rightarrow H_2O\)

b) \(NH_4Cl+CsOH\rightarrow CsCl+NH_3\uparrow+H_2O\)

- Pt ion đầy đủ: \(NH_4^++Cl^-+Cs^++OH^-\rightarrow Cs^++Cl^-+NH_3\uparrow+H_2O\)

- Pt ion rút gọn: \(NH_4^++OH^-\rightarrow NH_3\uparrow+H_2O\)

c) \(3Cu+8HNO_3\rightarrow3Cu\left(NO_3\right)_2+2NO\uparrow+4H_2O\)

- Pt ion đầy đủ: \(3Cu+8H^++8NO_3^-\rightarrow3Cu^{2+}+6NO_3^-+2NO\uparrow+4H_2O\)

- Pt ion rút gọn: \(3Cu+8H^++2NO_3^-\rightarrow3Cu^{2+}+2NO\uparrow+4H_2O\)

d) \(3HCl+Na_3BO_3\rightarrow3NaCl+H_3BO_3\)

- Pt ion đầy đủ: \(3H^++3Cl^-+3Na^++BO_3^{3-}\rightarrow3Na^++3Cl^-+H_3BO_3\)

- Pt ion rút gọn: \(3H^++BO_3^{3-}\rightarrow H_3BO_3\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: \(HgO+H_2\underrightarrow{t^o}Hg+H_2O\) (1)

            \(PbO+H_2\underrightarrow{t^o}PbO+H_2O\) (2)

            \(FeO+H_2\underrightarrow{t^o}Fe+H_2O\) (3)

            \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\) (4)

            \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\) (5) 

            \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\) (6)

a) \(n_{Hg\left(1\right)}=0,3\left(mol\right)\Rightarrow m_{Hg\left(1\right)}=0,3.201=60,3\left(g\right)\)

    \(n_{Pb\left(2\right)}=0,3\left(mol\right)\Rightarrow m_{Pb\left(2\right)}=0,3.207=62,1\left(g\right)\)

   \(n_{Fe\left(3\right)}=0,3\left(mol\right)\Rightarrow m_{Fe\left(3\right)}=0,3.56=16,8\left(g\right)\)

   \(n_{Fe\left(4\right)}=0,2\left(mol\right)\Rightarrow m_{Fe\left(4\right)}=0,2.56=11,2\left(g\right)\)

   \(n_{Cu\left(5\right)}=0,3\left(mol\right)\Rightarrow m_{Cu\left(5\right)}=0,3.64=19,2\left(g\right)\)

   \(n_{Fe\left(6\right)}=0,225\left(mol\right)\Rightarrow m_{Fe\left(6\right)}=0,225.56=12,6\left(g\right)\)

b)

\(n_{HgO\left(1\right)}=0,3\left(mol\right)\Rightarrow m_{HgO\left(1\right)}=0,3.217=65,1\left(g\right)\)

\(n_{PbO\left(2\right)}=0,3\left(mol\right)\Rightarrow m_{PbO\left(2\right)}=0,3.223=66,9\left(g\right)\)

\(n_{FeO\left(3\right)}=0,3\left(mol\right)\Rightarrow m_{FeO\left(3\right)}=0,3.72=21,6\left(g\right)\)

\(n_{Fe_2O_3\left(4\right)}=0,1\left(mol\right)\Rightarrow m_{Fe_2O_3\left(4\right)}=0,1.160=16\left(g\right)\)

\(n_{CuO\left(5\right)}=0,3\left(mol\right)\Rightarrow m_{CuO\left(5\right)}=0,3.80=24\left(g\right)\)

\(n_{Fe_3O_4\left(6\right)}=0,075\left(mol\right)\Rightarrow m_{Fe_3O_4}=0,075.232=17,4\left(g\right)\)

a) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

            0,2<----0,4<-----0,2<----0,2

\(\%m_{Mg}=\dfrac{0,2.24}{8,8}.100\%=54,5\%\)

\(\%m_{MgO}=100\%-54,5\%=45,5\%\)

b) \(n_{MgO}=\dfrac{8,8.45,5\%}{40}=0,1\left(mol\right)\)

PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

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

=> n_HCl = 0,4 + 0,2 = 0,6 (mol)

=> \(V_{dd.HCl}=\dfrac{0,6}{0,8}=0,75\left(l\right)\)

c) \(C_{M\left(MgCl_2\right)}=\dfrac{0,1+0,2}{0,75}=0,4M\)

Gọi số mol AgNO₃ là x (mol)

\(\left\{{}\begin{matrix}n_{KCl}=0,1.0,1=0,01\left(mol\right)\\n_{KI}=0,1.0,1=0,01\left(mol\right)\end{matrix}\right.\)

- Nếu kết tủa chỉ có AgI

\(n_{AgI}=\dfrac{3,211}{235}=0,01366\left(mol\right)>0,01\left(mol\right)\)

=> Vô lí

=> Kết tủa gồm AgI và AgCl

PTHH: \(AgNO_3+KI\rightarrow AgI\downarrow+KNO_3\)

               0,01<----0,01->0,01

\(\Rightarrow n_{AgCl}=\dfrac{3,211-0,01.235}{143,5}=0,006\left(mol\right)\)

PTHH: \(AgNO_3+KCl\rightarrow AgCl\downarrow+KNO_3\)

                0,006<----------0,006

=> \(V_{dd.AgNO_3}=\dfrac{0,006+0,01}{0,1}=0,16\left(l\right)\)

\(n_{KI}=n_{KCl}=0,1.0,1=0,01\left(mol\right)\)

PTHH:

\(KI+AgNO_3\rightarrow AgI\downarrow+KNO_3\)

0,01->0,01------>0,01

\(\rightarrow m_{AgI}=0,01.235=2,35\left(g\right)\\ \rightarrow m_{AgCl}=3,211-2,35=0,861\left(g\right)\\ \rightarrow n_{AgCl}=\dfrac{0,861}{143,5}=0,006\left(mol\right)\)

PTHH: \(KCl+AgNO_3\rightarrow AgCl\downarrow+KNO_3\)

                          0,006<---0,006

\(\rightarrow V_{dd.AgNO_3}=\dfrac{0,01+0,006}{0,1}=0,16M\)

Các ion phản ứng được với nhau là:

\(Ca^{2+}+CO_3^{2-}\rightarrow CaCO_3\downarrow\)

\(Ba^{2+}+CO_3^{2-}\rightarrow BaCO_3\downarrow\)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)

\(Fe^{3+}+3OH^-\rightarrow Fe\left(OH\right)_3\downarrow\)

\(2Fe^{3+}+3CO_3^{2-}+3H_2O\rightarrow2Fe\left(OH\right)_3\downarrow+3CO_2\uparrow\)

\(Fe^{3+}+3HCO_3^-\rightarrow Fe\left(OH\right)_3\downarrow+3CO_2\uparrow\)

\(NH_4^++OH^-\rightarrow NH_3\uparrow+H_2O\)

\(OH^-+HCO_3^-\rightarrow CO_3^{2-}+H_2O\)

a) Hằng số cân bằng: \(K_{cb}=\dfrac{\left[NH_3\right]^2}{\left[N_2\right]\cdot\left[H_2\right]^3}=\dfrac{2^2}{1\cdot3^3}=\dfrac{4}{27}\)

b) Thêm 1 mol N₂, cân bằng sẽ dịch chuyển theo chiều thuận.

\(N_2+3H_2⇌2NH_3\)

\(\left(2\right)\rightarrow3\) ..... \(\rightarrow2\)

\(x\rightarrow3x\) ..... \(\rightarrow2x\)

CB: \(2-x\rightarrow3-3x\rightarrow2+2x\)

Hằng số cân bằng luôn không đổi, nên ta có:\(K_{cb}=\dfrac{\left(2+2x\right)^2}{\left(2-x\right)\left(3-3x\right)^3}=\dfrac{4}{27}\)

\(\Leftrightarrow x\approx0,124\)

\(\Rightarrow\left\{{}\begin{matrix}\left[N_2\right]=2-0,124=1,876\left(M\right)\\\left[H_2\right]=3-3\cdot0,124=2,628\left(M\right)\\\left[NH_3\right]=2+2\cdot0,124=2,248\left(M\right)\end{matrix}\right.\).

`Fe + H_2 SO_4 -> FeSO_4 + H_2 \uparrow`

`0,2`      `0,2`                 `0,2`                         `(mol)`

`n_[Fe]=[11,2]/56=0,2(mol)`

`n_[H_2 SO_4]=0,4.1=0,4(mol)`

Có: `[0,2]/1 < [0,4]/1 ->H_2 SO_4` dư

 `=>m_[FeSO_4]=0,2.152=30,4(g)`

\(n_{Fe}=0,2 \left(mol\right), n_{H_2SO_4}=0,4 \left(mol\right)\)

pthh: \(2Fe+6H_2SO_4đ\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

.......... \(\dfrac{2}{15}\leftarrow0,4\) ............ \(\rightarrow\dfrac{1}{15}\)

(Fe dư \(\dfrac{1}{15}\) mol)

\(\Rightarrow Fe+Fe_2\left(SO_4\right)_3\rightarrow3FeSO_4\)

.... \(\dfrac{1}{15}\rightarrow\dfrac{1}{15}\) ............ \(\rightarrow\dfrac{3}{15}\)

\(m_{\text{muối}}=\dfrac{3}{15}\cdot152=30,4\) (gam)

Do nguyên tử X có tổng số hạt là 36

=> 2p + n = 36 (1)

Do số hạt không mang điện bằng một nửa hiệu số giữa tổng số hạt với số hạt mang điện tích âm

=> \(n=\dfrac{1}{2}\left(36-e\right)=\dfrac{1}{2}\left(36-p\right)\) (2)

(1)(2) => p = 12 (hạt);  n = 12 (hạt)

=> X là Mg

\(n_{CuO}=\dfrac{80}{80}=1\left(mol\right)\)

PTHH: 2Cu + O₂ --t^o--> 2CuO

              1<---0,5<--------1

\(\left\{{}\begin{matrix}m_{Cu}=1.64=64\left(g\right)\\m_{O_2}=0,5.32=16\left(g\right)\end{matrix}\right.\)

a)

Gọi số mol Fe phản ứng là a (mol)

PTHH: Fe + CuSO₄ --> FeSO₄ + Cu

             a------>a---------->a----->a

=> m_tăng = 64a - 56a = 2,4 (g)

=> a = 0,3 (mol)

=> m_Fe(pư) = 0,3.56 = 16,8 (g)

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

dd sau pư chứa \(\left\{{}\begin{matrix}CuSO_4:0,6-0,3=0,3\left(mol\right)\\FeSO_4:0,3\left(mol\right)\end{matrix}\right.\)

PTHH: \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

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

            \(FeSO_4+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+Na_2SO_4\)

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

             \(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

                    0,3----->0,3

              \(4Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)

                     0,3------------>0,15

=> x = 0,3.80 + 0,15.160 = 48 (g)

\(Fe+CuSO_4\rightarrow FeSO_4+Cu\\ n_{CuSO_4}=0,3.2=0,6\left(mol\right)\\ Đặt:n_{Fe\left(pứ\right)}=x\left(mol\right)\\ Tacó:64x-56x=2,4\\ \Rightarrow x=0,3\left(mol\right)\\ \Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\\ b.Dungdịchthuđược:\\ FeSO_4:0,3\left(mol\right);CuSO_4dư:0,6-0,3=0,3\left(mol\right)\\ Chấtrắnthuđược:CuO,Fe_2O_3\\ BảotoànnguyêntốCu:n_{CuSO_4}=n_{CuO}=0,3\left(mol\right)\\ BảotoànnguyêntốFe:n_{FeSO_4}=n_{Fe_2O_3}.2\\ \Rightarrow n_{Fe_2O_3}=0,15\left(mol\right)\\ x=m_{Fe_2O_3}+m_{CuO}=0,3.80+0,15.160=48\left(g\right)\)