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 $L$.

$m_{\ce{CO2(aq)}} = [\ce{CO2(aq)}]v =  \frac{p_{\ce{CO2(gas)atm}}}{K_{AW}}v$