color(red)("Warning! Long answer!"
All the credit for this answer goes to Stefan V.
The big clues are the yellow flame colour ("Na") for bb"G" and the brown gas obtained from bb"B" ("NO"_2 or, less likely, "Br"_2).
"NO"_2 comes from the decomposition of "HNO"_2.
This suggests that bb"B" is "NaNO"_2 and that bb"A" is probably "NaNO"_3.
Acting on that assumption, we get
2underbrace("NaNO"_3)_color(red)("A") stackrelcolor(blue)(Δcolor(white)(X)) (→) 2underbrace("NaNO"_2)_color(red)("B") + underbrace("O"_2)_color(red)("C")
Then gas color(red)(bb"C"color(white)(l) "is"color(white)(l) "O"_2).
"O"_2 is a colourless gas that forms "MgO" when heated with "Mg".
"O"_2 + "2Mg" stackrelcolor(blue)(Δcolor(white)(m))(→) underbrace("2MgO")_color(red)("white powder")
"MgO" is a basic oxide.
"MgO + H"_2"O" ⇌ "Mg"^(2+) + "2OH"^(-)
color(red)(bb"B" = "NaNO"_2)
"NaNO"_2 reacts with sulfuric acid to form "HNO"_2, which decomposes to form the brown gas "NO"_2.
2underbrace("NaNO"_2)_color(red)("B") + "H"_2"SO"_4 → "2HNO"_2 + "Na"_2"SO"_4
"2HNO"_2 → underbrace("NO"_2)_color(red)("brown gas") + "NO" + "H"_2"O"
color(red)(bb"D" = "N"_2) and color(red)(bb"E" color(white)(l) = color(white)(l)"NaCl")
"NaNO"_2 reacts with "NH"_4"Cl" to form "N"_2 and "NaCl".
underbrace("NaNO"_2)_color(red)("B") + "NH"_4"Cl" → underbrace("N"_2)_color(red)("D") + underbrace("NaCl")_color(red)("E") + "2H"_2"O"
"N"_2 reacts with "Ca" to form "Ca"_3"N"_2.
"3Ca" + "N"_2 stackrelcolor(blue)(Δcolor(white)(X))(→) "Ca"_3"N"_2
"Ca"_3"N"_2 reacts with water to form ammonia and "Ca"("OH")_2.
"Ca"_3"N"_2 + 6"H"_2"O" → 3underbrace("Ca"("OH")_2)_color(red)("basic") + 2underbrace("NH"_3)_color(red)("colourless")
color(red)(bb"F" = "N"_2"O") and color(red)(bb"G" = "Na"_2"SO"_4")
"NaNO"_3 reacts with ("NH"_4)_2"SO"_4 to form "N"_2"O" and "Na"_2"SO"_4
2underbrace("NaNO"_3)_color(red)("A") + ("NH"_4)_2"SO"_4 stackrelcolor(blue)(Δcolor(white)(X))(→) 2underbrace("N"_2"O")_color(red)("F") + underbrace("Na"_2"SO"_4)_color(red)("G") + "4H"_2"O"
The "Na" in "Na"_2"SO"_4 gives the yellow flame colour.
Formula Units of bb"A"
You don't give a pressure or temperature for the gas bb"D", so I will arbitrarily assume that P = "1 atm" and T = "25 °C".
Then, n =(PV)/(RT) = (1 color(red)(cancel(color(black)("atm"))) ×0.120 color(red)(cancel(color(black)("L"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" ×298.15 color(red)(cancel(color(black)("K")))) = 4.90 × 10^"-3" color(white)(l)"mol"
"Moles of NaNO"_3 = 4.90 × 10^"-3"color(red)(cancel(color(black)("mol N"_2))) × (1 color(red)(cancel(color(black)("mol NaNO"_2))))/(1 color(red)(cancel(color(black)("mol N"_2)))) × ("2 mol NaNO"_3)/(2 color(red)(cancel(color(black)("mol NaNO")))_2) = 4.90 × 10^"-3" "mol NaNO"_3
"Formula units NaNO"_3 = 4.90 × 10^"-3" color(red)(cancel(color(black)("mol NaNO"_3))) × (6.022 × 10^23 "formula units NaNO"_3)/(1 color(red)(cancel(color(black)("mol NaNO"_3)))) = 2.95 ×10^21color(white)(l) "formula units NaNO"_3
