\(\%K=\dfrac{m_K}{M_{K_2SO_3}}=\dfrac{78}{158}=49,36\%\\ \%S=\dfrac{m_S}{M_{K_2SO_3}}=\dfrac{32}{158}=20,25\%\\ \%O=100\%-\%K-\%S=100\%-49,36\%-20,25\%=30,39\%\)

cảm ơn ạ

\(d_{\dfrac{A}{H_2}}=8,5\\ M_{H_2}=2\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_A=d_{\dfrac{A}{H_2}}.M_{H_2}=8,5.2=17\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow m_N=\%N.M_A=82,35\%.17=14\left(g\right)\\ m_H=m_A-m_N=17-14=3\left(g\right)\\ \Rightarrow n_N=\dfrac{m}{M}=\dfrac{14}{14}=1\left(mol\right)\\ n_H=\dfrac{m}{M}=\dfrac{3}{1}=3\left(mol\right)\\ CTHH:NH_3\)

\(M_A=M_{H_2}.8,5=2.8,5=17(g/mol)\)

Đặt CTHH của \(A\) là \(N_xH_y\)

Ta có : \(\dfrac{14x.100}{17}=82,35\)

\(-> x=1\)

Ta có : \(\dfrac{y.100}{17}=17,65\)

\(->y=3\)

Vậy CTHH của \(A\) là \(NH_3\)

b

\(a.PTHH:3Fe+2O_2\overset{t^o}{--->}Fe_3O_4\)

b. \(m_{Fe}+m_{O_2}=m_{Fe_3O_4}\)

c. Thay các đại lượng vào b, ta được:

\(25,2+m_{O_2}=34,8\)

\(\Leftrightarrow m_{O_2}=34,8-25,2=9,6\left(g\right)\)

Ta có: \(n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)

\(PTHH:2Al+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2\uparrow\)

               0,4 <---  0,6    ----------->      0,2    -->   0,6

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4.27=10,8\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\\V_{H_2}=0,6.22,4=13,44\left(lít\right)\end{matrix}\right.\)

\(a, 2Al+3Cl_2->2AlCl_3\)

\(b,2KClO_3->2KCl+3O_2\)

\(c, 2H_3PO_4+3Ba(OH)_2->Ba_3(PO_4)_2+6H_2O\)

\(\left(a\right)2Al+3Cl_2\overset{t^o}{--->}2AlCl_3\)

\(\left(b\right)2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\uparrow\)

\(\left(c\right)2NaOH+CuSO_4--->Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

\(\left(d\right)2H_3PO_4+3Ba\left(OH\right)_2--->Ba_3\left(PO_4\right)_2+6H_2O\)

\(a,PTHH:Zn+HCl\rightarrow ZnCl_2+H_2\)

\(b,Áp.dụng.ĐLBTKL,ta.có:\)

\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ m_{H_2}=m_{Zn}+m_{HCl}-m_{ZnCl_2}=9,75+10,95-20,4=0,3\left(g\right)\)

cảm ơn

\(\%Fe=\dfrac{m_{Fe}}{M_{Fe_2O_3}}=\dfrac{112}{160}=70\%\\ \%O=100\%-\%Fe=100\%-70\%=30\%\)

cảm ơn

\(a,2KHCO_3\rightarrow\left(t^o\right)K_2CO_3+CO_2+H_2O\\ b.n_{K_2CO_3}=\dfrac{27,6}{138}=0,2\left(mol\right)\\ n_{KHCO_3}=2.0,2=0,4\left(mol\right)\\ m_{KHCO_3}=100.0,4=40\left(g\right)\\ c,n_{CO_2}=n_{H_2O}=n_{K_2CO_3}=0,2\left(mol\right)\\ V_{hh\left(CO_2,H_2O\right)}=\left(0,2+0,2\right).22,4=8,96\left(l\right)\)

a, PTHH: 2KHCO₃ -t^o-> K₂CO₃ + H₂O + CO₂

b, n_K2CO3 = m/M = 27,6/138 = 0,2 (mol)

Theo PTHH: n_KHCO3 = 2.n_K2CO3 = 2 . 0,2 = 0,4 (mol)

=> m_KHCO3 = n.M = 0,4 . 100 = 40 (g)

c, Theo PTHH: n_CO2 = n_H2O = n_K2CO3 = 0,2 (mol)

=> n_hh = n_CO2 + n_H2O = 0,2 + 0,2 = 0,4 (mol)

Ta có: Thể tích khí ở 20^oC và 1 atm là đkt

=> V_hh(đkt) = n.24 = 0,4 . 24 = 9,6 (l)