# Question f22b3

Apr 23, 2016

#### Explanation:

color(red)("Warning! Long answer!"

All the credit for this answer goes to Stefan V.

The big clues are the yellow flame colour ($\text{Na}$) for $\boldsymbol{\text{G}}$ and the brown gas obtained from $\boldsymbol{\text{B}}$ (${\text{NO}}_{2}$ or, less likely, ${\text{Br}}_{2}$).

${\text{NO}}_{2}$ comes from the decomposition of ${\text{HNO}}_{2}$.

This suggests that $\boldsymbol{\text{B}}$ is ${\text{NaNO}}_{2}$ and that $\boldsymbol{\text{A}}$ is probably ${\text{NaNO}}_{3}$.

Acting on that assumption, we get

$2 \underbrace{\text{NaNO"_3)_color(red)("A") stackrelcolor(blue)(Δcolor(white)(X)) (→) 2underbrace("NaNO"_2)_color(red)("B") + underbrace("O"_2)_color(red)("C}}$

Then gas $\textcolor{red}{{\boldsymbol{\text{C"color(white)(l) "is"color(white)(l) "O}}}_{2}}$.

${\text{O}}_{2}$ is a colourless gas that forms $\text{MgO}$ when heated with $\text{Mg}$.

"O"_2 + "2Mg" stackrelcolor(blue)(Δcolor(white)(m))(→) underbrace("2MgO")_color(red)("white powder")

$\text{MgO}$ is a basic oxide.

${\text{MgO + H"_2"O" ⇌ "Mg"^(2+) + "2OH}}^{-}$

$\textcolor{red}{{\boldsymbol{\text{B" = "NaNO}}}_{2}}$

${\text{NaNO}}_{2}$ reacts with sulfuric acid to form ${\text{HNO}}_{2}$, which decomposes to form the brown gas ${\text{NO}}_{2}$.

2underbrace("NaNO"_2)_color(red)("B") + "H"_2"SO"_4 → "2HNO"_2 + "Na"_2"SO"_4

$\text{2HNO"_2 → underbrace("NO"_2)_color(red)("brown gas") + "NO" + "H"_2"O}$

$\textcolor{red}{{\boldsymbol{\text{D" = "N}}}_{2}}$ and $\textcolor{red}{\boldsymbol{\text{E" color(white)(l) = color(white)(l)"NaCl}}}$

${\text{NaNO}}_{2}$ reacts with $\text{NH"_4"Cl}$ to form ${\text{N}}_{2}$ and $\text{NaCl}$.

underbrace("NaNO"_2)_color(red)("B") + "NH"_4"Cl" → underbrace("N"_2)_color(red)("D") + underbrace("NaCl")_color(red)("E") + "2H"_2"O"

${\text{N}}_{2}$ reacts with $\text{Ca}$ to form ${\text{Ca"_3"N}}_{2}$.

${\text{3Ca" + "N"_2 stackrelcolor(blue)(Δcolor(white)(X))(→) "Ca"_3"N}}_{2}$

${\text{Ca"_3"N}}_{2}$ 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")

$\textcolor{red}{\boldsymbol{\text{F" = "N"_2"O}}}$ and $\textcolor{red}{\boldsymbol{\text{G" = "Na"_2"SO"_4}}}$

${\text{NaNO}}_{3}$ reacts with ("NH"_4)_2"SO"_4 to form $\text{N"_2"O}$ and ${\text{Na"_2"SO}}_{4}$

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 $\text{Na}$ in ${\text{Na"_2"SO}}_{4}$ gives the yellow flame colour.

Formula Units of $\boldsymbol{\text{A}}$

You don't give a pressure or temperature for the gas $\boldsymbol{\text{D}}$, so I will arbitrarily assume that $P = \text{1 atm}$ and $T = \text{25 °C}$.

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"#

${\text{Moles of NaNO"_3 = 4.90 × 10^"-3"color(red)(cancel(color(black)("mol N"_2))) × (1 color(red)(cancel(color(black)("mol NaNO"_2))))/(1 color(red)(cancel(color(black)("mol N"_2)))) × ("2 mol NaNO"_3)/(2 color(red)(cancel(color(black)("mol NaNO")))_2) = 4.90 × 10^"-3" "mol NaNO}}_{3}$

${\text{Formula units NaNO"_3 = 4.90 × 10^"-3" color(red)(cancel(color(black)("mol NaNO"_3))) × (6.022 × 10^23 "formula units NaNO"_3)/(1 color(red)(cancel(color(black)("mol NaNO"_3)))) = 2.95 ×10^21color(white)(l) "formula units NaNO}}_{3}$