\(n_{MgCl_2}=0,15.1=0,15\left(mol\right);n_{NaOH}=\dfrac{80.20\%}{40}=0,4\left(mol\right)\)

PTHH:         \(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)

Ban đầu:      0,15           0,4

Pư:              0,15------->0,3

Sau pư:       0                0,1               0,15

                   \(Mg\left(OH\right)_2\xrightarrow[]{t^o}MgO+H_2O\)

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

=> \(m=m_{MgO}=0,15.40=6\left(g\right)\)

`Na_2O  + H_2O -> 2NaOH`

`2Na + 2H_2O -> 2NaOH + H_2`

`CaCO_3 + 2HCl -> CaCl_2 + CO_2 + H_2O`

`CO_2 + Ca(OH)_2 -> CaCO_3 + H_2O`

`Fe + 2HCl -> FeCl_2 + H_2`

a) \(n_M=\dfrac{4,2}{M_M}\left(mol\right);n_{M_xO_y}=\dfrac{5,8}{xM_M+16y}\left(mol\right)\)

PTHH: \(2xM+yO_2\xrightarrow[]{t^o}2M_xO_y\)

Theo PT: \(n_M=x.n_{M_xO_y}\)

=> \(\dfrac{4,2}{M_M}=x.\dfrac{5,8}{xM_M+16y}\)

=> \(M_M=\dfrac{42y}{x}\left(g/mol\right)=21.\dfrac{2y}{x}\left(g/mol\right)\)

Chỉ có \(\dfrac{2y}{x}=\dfrac{8}{3}\) thỏa mãn => \(\dfrac{x}{y}=\dfrac{3}{4}\)

=> \(M_M=21.\dfrac{8}{3}=56\left(g/mol\right)\)

=> M là Fe, oxit là Fe₃O₄

`Fe + 2HCl -> FeCl_2 + H_2`

`2Na + 2HCl -> 2NaCl + H_2`

`2Al + 6HCl -> 2AlCl_3 + 3H_2`

`2Al(OH)_3 + 3H_2SO_4 -> Al_2(SO_4)_3 + 6H_2O`

Vậy phần nước là dung môi đi đâu rồi???

a) \(n_{Na}=\dfrac{3,45}{23}=0,15\left(mol\right)\)

PTHH: 2Na + 2H₂O ---> 2NaOH + H₂ 

           0,15---------------->0,15---->0,075

=> \(m_{\text{dd}NaOH}=\dfrac{0,15.40}{10\%}=60\left(g\right)\)

Ta có: \(m_{\text{dd}NaOH}=m_{Na}+m_{H_2O}-m_{H_2}\)

=> \(m=m_{H_2O}=60-3,45+0,075.2=56,7\left(g\right)\)

b) \(n_{O_2}=\dfrac{1,6}{32}=0,05\left(mol\right)\)

PTHH: \(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

Xét tỉ lệ: \(\dfrac{0,075}{2}< \dfrac{0,05}{1}\Rightarrow O_2\) dư, H₂ không đủ để đốt cháy hết

Theo PTHH: \(n_{H_2O}=n_{H_2}=0,075\left(mol\right)\)

=> \(m_{s\text{ản}.ph\text{ẩm}}=m_{H_2O}=0,075.18=1,35\left(g\right)\)

\(n_{SO_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ \Rightarrow m_{SO_2}=n.M=0,1.64=6,4\left(g\right)\\ n_{O_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ \Rightarrow m_{O_2}=n.M=0,05.32=1,6\left(g\right)\\ \Rightarrow m_{hh}=m_{SO_2}+m_{O_2}=6,4+1,6=8\left(g\right)\)

Câu 1: D

Câu 2: A

Câu 3: C

Câu 4: D

`1)`

`a) 3Fe + 2O_2 -> Fe_3O_4`

`b) 3NaOH + FeCl_3 -> Fe(OH)_3 + 3NaCl`

`c) N_2 + 3H_2 -> 2NH_3`

`d) 2KClO_3 -> 2KCl + 3O_2` (Đáng ra là `KClO_3` chứ?)

`e) 4C_xH_y + (4x+y)O_2 -> 4xCO_2 + 2yH_2O`

`2)`

`a) 2Mg + O_2 -> 2MgO`

`b) 2Fe + 3Cl_2 -> 2FeCl_3`

`c) 2NaOH + CuCl_2 -> Cu(OH)_2 + 2NaCl`

`d) H_2 + Cl_2 -> 2HCl`

`e) Mg + 2HCl -> MgCl_2 + H_2`

`3)`

`a) n_{Fe} = m/M = (5,6)/(56) = 0,1 (mol)`

`b) V_{H_2(đktc)} = n.22,4 = 0,25.22,4 = 5,6 (l)`

`c) n_{CO_2} = (V_{(đktc)})/(22,4) = (4,48)/(22,4) = 0,2 (mol)`

`=> m_{CO_2} = n.M = 0,2.44 = 8,8 (g)`

`d)` \(n_{CO_2}=\dfrac{9,6.10^{23}}{6.10^{23}}=1,6\left(mol\right)\)

`4)`

`a) CO_2: \%C = (12)/(44) .100\% = 27,27\%`

`\%H = 100\% - 27,27\% = 72,73\%`

`b) CaO: \%Ca = (40)/(56) .100\% = 71,43\%`

`\%O = 100\% - 71,43\% = 28,57\%`

a) \(n_{CO_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

`=>` \(m_{CO_2}=n.M=0,6.44=26,4\left(g\right)\)

b) \(n_{SO_2}=\dfrac{m}{M}=\dfrac{12,8}{64}=0,2\left(mol\right)\)

`=>` \(V_{SO_2\left(\text{đ}ktc\right)}=n.22,4=0,2.22,4=4,48\left(l\right)\)