Antacids, such as alka-seltzer, use the reaction of sodium bicarbonate with citric acid in water solution to produce a fizz as follows: 3nahco3 + c6h8o7 → 3co2 + 3h2o + na3c6h5o7 if 4.11 g of the citric acid (c6h8o7, mw = 192 g/mol) react with excess sodium bicarbonate (nahco3), how many grams of carbon dioxide (co2, mw = 44 g/mol) are formed as the solution fizzes?

2.8248 g

Explanation:

First, consider the balanced chemical equation:

Then we calculate the number of moles in 4.11 g of citric acid:

n(citric acid)=

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

the answer to your question is c

lol for the pointsss