Câu 1:

$a,2N_2+3H_2\xrightarrow{t^o,p,xt}2NH_3\\ b,BaO+2HNO_3\to Ba(NO_3)_2+H_2O\\ c,Ca+2H_2O\to Ca(OH)_2+H_2\\ d,2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3\\ e,4P+5O_2\xrightarrow{t^o}2P_2O_5\\ f,Zn(OH)_2+2HCl\to ZnCl_2+2H_2O\\ g,3Mg+2H_3PO_4\to Mg_3(PO_4)_2+3H_2O\\ h,C_2H_4+3O_2\xrightarrow{t^o}2CO_2+2H_2O$

Câu 2:

$a,n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1(mol)\\ b,m_{SO_2}=0,6.64=38,4(g)\\ c,V_{O_2(đkc)}=1,8.22,4=40,32(l)$

Câu 3:

$a,PTHH:4K+O_2\xrightarrow{t^o}2K_2O\\ \text{Tỉ lệ: }4:1:2\\ b,\text{Bảo toàn KL: }m_K+m_{O_2}=m_{K_2O}$

Em tham khảo nha :

\(\begin{array}{l}
1)\\
{N_2} + 3{H_2} \to 2N{H_3}\\
BaO + 2HN{O_3} \to Ba{(N{O_3})_2} + {H_2}O\\
Ca + 2{H_2}O \to Ca{(OH)_2} + {H_2}\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
2P + 5Cl \to 2PC{l_5}\\
Zn{(OH)_2} + 2HCl \to ZnC{l_2} + 2{H_2}O\\
3Mg + 2{H_3}P{O_4} \to M{g_3}{(P{O_4})_2} + 3{H_2}\\
{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O
\end{array}\)

\(\begin{array}{l}
2)\\
{n_{N{a_2}C{O_3}}} = \dfrac{{10,6}}{{106}} = 0,1mol\\
{m_{S{O_2}}} = 0,6 \times 64 = 38,4g\\
{V_{{O_2}}} = 1,8 \times 24,79 = 44,622l\\
3)\\
a)\\
4K + {O_2} \to 2{K_2}O\\
{N_K}:{N_{{O_2}}}:{N_{{K_2}O}} = 4:1:2\\
b)\\
BTKL:\\
{m_K} + {m_{{O_2}}} = {m_{{K_2}O}}\\
 \Rightarrow {m_{{O_2}}} = 188 - 156 = 32g
\end{array}\)