Number of grams of CO2 released from a carbonated water

The partial pressure of $\ce{CO2}$ gas in a bottle of carbonated water is $3.84\mathrm{~atm}$ at $25\mathrm{~^\circ C}$. How much $\ce{CO2}$ gas (in g) will be released from $0.5\mathrm{~L}$ of the carbonated water when the partial pressure of $\ce{CO2}$ is lowered to $1.04\mathrm{~atm}$? $25\mathrm{~^\circ C}$, the Henry’s law constant for $\ce{CO2}$ dissolved in water is $1.65 \times 10^3\mathrm{~atm}$, and the density of water is $1.0\mathrm{~g/cm^3}$.


First, I found the number of moles of $\ce{CO2}$ using this formula:


$$n = \frac{PV}{k}$$


So the answer that I found which is $0.000351$, can I multiply it by a molar mass of a $\ce{CO2}$ to get the number of grams inside the bottle? This approach seems too easy for me, but at the same time it makes sense. 




This shows **part** of the problem. 


Henry's law as given is 


$K\_{AW}^{px} = $1.65\*10^3$ = \frac{p\_{\ce{CO2(gas)atm}}}{[\ce{CO2(aq)}]}$


where:
$K\_{AW}^{px}$ atm = aqueous-phase mixing ratio of $\ce{CO2}$ and the units are $atm$


$[\ce{CO2{(aq)}}]$ = the concentration of $\ce{CO2{(aq)}}$ in $mole/L$


$p\_{\ce{CO2(gas)atm}}]$ = the partial pressure of $\ce{CO2{(gas)}}$ in $atm$ 


So rearranging...


$[\ce{CO2(aq)}] =\frac{p\_{\ce{CO2(gas)atm}}}{K\_{AW}}$


The moles of $\ce{CO2}$, $m\_{\ce{CO2}}$, is given by concentration, $[\ce{CO2}]$, times the volume, $v$ in $L$.


$m\_{\ce{CO2(aq)}} = [\ce{CO2(aq)}]v = \frac{p\_{\ce{CO2(gas)atm}}}{K\_{AW}}v$


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