Answer:
2.8248 g
Explanation:
First, consider the balanced chemical equation:
[tex]3NaHCO_{3} +C_{6}H_{8}O_{7} --->3CO_{2} +3H_{2}O +Na_{3}C_{6}H_{5}O_{7}\\[/tex]
Then we calculate the number of moles in 4.11 g of citric acid:
n(citric acid)=[tex]\frac{4.11g}{192g/mol}=0.0214mol[/tex]
According to the balanced reaction, one mole of citric acid produces 3 moles of carbon dioxide. That's 3 times the number f moles of citric acid. So we will do the same with the available number of moles of citric acid.
so n(carbon dioxide) = 0.0214 mol*3=0.0642 mol
mass(carbon dioxide)= mass*molar mass=0.0642 mol* 44g/mol
= 2.8248 g