# Question badf7

Feb 2, 2017

$\textsf{\left(a\right)}$

$\textsf{{p}_{C {O}_{2}} = {p}_{{H}_{2} O} = 0.50 \textcolor{w h i t e}{x} \text{Atm}}$

$\textsf{\left(b\right)}$

$\textsf{{m}_{N {a}_{2} C {O}_{3}} = 1.62 \textcolor{w h i t e}{x} \text{g}}$

$\textsf{{m}_{N a H C {O}_{3}} = 11.0 \textcolor{w h i t e}{x} g}$

$\textsf{\left(c\right)}$

$\textsf{V = 5.28 \textcolor{w h i t e}{x} L}$

#### Explanation:

$\textsf{\left(a\right)}$

$\textsf{2 N a H C {O}_{3 \left(s\right)} \rightarrow N {a}_{2} C {O}_{3 \left(s\right)} + C {O}_{2 \left(g\right)} + {H}_{2} {O}_{\left(g\right)}}$

$\textsf{{K}_{p} = {p}_{C {O}_{2}} \times {p}_{{H}_{2} O} = 0.25 \textcolor{w h i t e}{x} {\text{Atm}}^{2}}$

Since sf(p_(CO_2)=p_(H_2O)# we can say:

$\textsf{{p}_{C {O}_{2}}^{2} = 0.25}$

$\therefore$$\textsf{{p}_{C {O}_{2}} = \sqrt{0.25} = \textcolor{red}{0.50 \textcolor{w h i t e}{x} \text{Atm}}}$

and

$\textsf{{p}_{{H}_{2} O} = \textcolor{red}{0.50 \textcolor{w h i t e}{x} \text{Atm}}}$

$\textsf{\left(b\right)}$

We can get the initial moles $\textsf{n}$ of $\textsf{N a H C {O}_{3}}$ using:

$\textsf{n = \frac{m}{M} _ r = \frac{13.6}{84.0} = 0.162}$

Set up an ICE table based on moles:

$\textsf{\text{ } 2 N a H C {O}_{3 \left(g\right)} \rightarrow N {a}_{2} C {O}_{3 \left(s\right)} + C {O}_{2 \left(g\right)} + {H}_{2} {O}_{\left(g\right)}}$

$\textsf{I \text{ "0.162" "0" "0" } 0}$

$\textsf{C \text{ "-2x" "+x" "+x" } + x}$

$\textsf{E \text{ "(0.162-2x) " "x" "x" } x}$

We can find x from the volume of $\textsf{C {O}_{2}}$:

$\textsf{P V = {n}_{C {O}_{2}} R T}$

$\therefore$$\textsf{{n}_{C {O}_{2}} = \frac{P V}{R T} = \frac{0.50 \times 1.00}{0.082 \times 398} = 0.0153}$

$\therefore$$\textsf{x = 0.0153}$

$\therefore$$\textsf{{n}_{N {a}_{2} C {O}_{3}} = 0.0153}$

$\therefore$$\textsf{{m}_{N {a}_{2} C {O}_{3}} = {n}_{N {a}_{2} C {O}_{3}} \times {M}_{r} = 0.0153 \times 105.98 = \textcolor{red}{1.621 \textcolor{w h i t e}{x} g}}$

$\textsf{{n}_{N a H C {O}_{3}} = 0.162 - 2 \times \left(0.0153\right) = 0.1314}$

$\therefore$$\textsf{{m}_{N a H C {O}_{3}} = {n}_{N a H C {O}_{3}} \times {M}_{r} = 0.1314 \times 84.00 = \textcolor{red}{11.04 \textcolor{w h i t e}{x} g}}$

$\textsf{\left(c\right)}$

$\textsf{2 N a H C {O}_{3 \left(g\right)} \rightarrow N {a}_{2} C {O}_{3 \left(s\right)} + C {O}_{2 \left(g\right)} + {H}_{2} {O}_{\left(g\right)}}$

$\textsf{2 m o l \text{ "rarr" "1mol" "+1mol" } 1 m o l}$

Converting to grams $\textsf{\Rightarrow}$

$\textsf{2 \times 84.00 g \text{ "rarr" "1mol" } m o l}$

$\textsf{168 g \text{ "rarr" "1mol" } 1 m o l}$

If all the $\textsf{N a H C {O}_{3}}$ decomposes we can say that:

$\textsf{1 g \text{ "rarr" " 1/(168)mol" } \frac{1}{168} m o l}$

$\therefore$$\textsf{13.6 g \text{ "rarr" "13.6/(168)mol" } \frac{13.6}{168} m o l}$

$\textsf{13.6 g \text{ "rarr" "0.081mol" } 0.081 m o l}$

So the total moles of gas produced =$\textsf{2 \times 0.081 = 0.162}$

If the gases expand aqainst the atmosphere we can use the ideal gas expression:

$\textsf{P V = n R T}$

$\therefore$$\textsf{V = \frac{n R T}{P} = \frac{0.162 \times 0.082 \times 398}{1.00} = \textcolor{red}{5.28 \textcolor{w h i t e}{x} L}}$

This ignores the volume taken up by any solids.