38: 1l

39 K,Mg,Zn

40 SO3

$\text{Câu 38: 1 lít}$ $\\$ $\text{Giải thích:}$ $\\$ $n_{MgCO_3}$ = $\dfrac{m}{M}$ = $\dfrac{21}{84}$= $0,25 (mol)$ $\\$ $MgCO_3 + 2HCl → MgCl_2 + CO_2↑+H_2O$ $\\$ $\text{ 0,25 0,5 0,25 0,25 /mol}$ $\\$ $V_{dd_{HCl}}$ = $\dfrac{V}{C_M}$ = $\dfrac{0,5}{0,5}$ $= 1 (lít)$ $\\$ $\text{Câu 39: K, Mg, Zn}$ $\\$ $\text{Câu 40:}$ $SO_3$ $\\$ $\text{Giải thích:}$ $\\$ $M_{SO_3}$ $\text{= 32 + 16.3 = 80 (g/mol)}$ $\\$ $\text{%S}$ = $\dfrac{32.100}{80}$ $\text{= 40%}$ $\\$ $\text{%O = 100 - %S = 100 - 40 = 60% (1)}$ $\\$ $M_{Fe_2O_3}$ $\text{= 56.2 + 16.3 = 160 (g/mol)}$ $\\$ $\text{%Fe =}$ $\dfrac{56.2.100}{160}$ $\text{= 70%}$ $\\$ $\text{%O = 100 - %Fe = 100 - 70 = 30% (2)}$ $\\$ $M_{CuO}$ $\text{= 64 + 16 = 80 (g/mol)}$ $\\$ $\text{%Cu =}$ $\dfrac{64.100}{80}$ $\text{= 80%}$ $\\$ $\text{%O = 100 - %Cu = 100 - 80 = 20% (3)}$ $\\$ $M_{SO_2}$ $\text{= 32 + 16.2 = 64 (g/mol)}$ $\\$ $\text{%S =}$ $\dfrac{32.100}{64}$ $\text{= 50%}$ $\\$ $\text{%O = 100 - %S = 100 - 50 = 50% (4)}$ $\\$ $\text{Từ (1), (2), (3) và (4) suy ra}$ $SO_3$ $\text{là chất có phần trăm khối lượng oxi lớn nhất.}$