\(M_X=23.M_{H_2}=23.2=46(g/mol)\)

Trong 1 mol khí X: \(\begin{cases} n_N=\dfrac{46.30,43\%}{14}=1(mol)\\ n_O=\dfrac{46-14}{16}=2(mol) \end{cases}\)

Vậy CTHH là \(NO_2\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1(Mol)\\ a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{ZnCl_2}=n_{Zn}=0,1(mol)\\ \Rightarrow m_{ZnCl_2}=0,1.136=13,6(g)\\ b,n_{H_2}=n_{Zn}=0,1(mol)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24(l)\)

\(b,n_{CaCO_3}=\dfrac{m}{M}=\dfrac{50}{100}=0,5(mol)\\ \Rightarrow n_O=3n_{CaCO_3}=1,5(mol)\\ \Rightarrow \text{Số nguyên tử oxi là: }A_O=1,5.6.10^{23}=9.10^{23}\)

\(300(ml)=0,3(l)\\ n_{HCl}=1.0,3=0,3(mol);n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ \text{LTL: }\dfrac{n_{Fe}}{1}<\dfrac{n_{HCl}}{2}\Rightarrow HCl\text{ dư}\\ \Rightarrow n_{HCl(dư)}=0,3-0,1.2=0,1(mol)\\ \Rightarrow m_{HCl(dư)}=0,1.36,5=3,65(g)\\ b,n_{FeCl_2}=n_{Fe}=0,1(mol)\\ \Rightarrow \begin{cases} C_{M_{FeCl_2}}=\dfrac{0,1}{0,3}=0,33M\\ C_{M_{HCl(dư)}}=\dfrac{0,1}{0,3}=0,33M \end{cases}\)

Bảo toàn K: \(n_K=3n_{K_3PO_4}=0,6(mol)\)

\(\Rightarrow n_{K_3PO_4}=0,2(mol)\\ \Rightarrow m_{K_3PO_4}=0,2.212=42,4(g)\)

câu b lập tỉ lệ chất nào còn dư chứ e =,=