Answer
$2.04\text{ L $CO_2$}$
Work Step by Step
$4.00\text{ grams $CO_2$}*\frac{1\text{ mol $CO_2$}}{44.01\text{ grams $CO_2$}}*\frac{22.42 \text{ L}}{1\text{ mol $CO_2$}}=2.04\text{ L $CO_2$}$
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