how many grams of carbon dioxide are formed as the solution fizzes?

Question

if 2.57g of the citric acid (C6H8O7, MW=192g/mol) reacts with the excess sodium bicarbonate (NaHCO3), how many grams of CO2 (MW=44g/mol)....? background info: 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.

Answer 1

The citric acid molecular weight of 192 is much larger than the amount you used - 2.57g. So you only used a small fraction of a mole. Calculate the fraction - it should be roughly around 1% of a mole.

The formula says that for every mole of citric acid you are going to get 3 moles of CO2. The CO2 doesnt weigh much, only about 44 grams per mole. You are going to get three moles though. So for every 192 grams of citric acid you should get 44 x3 = 132 grams of CO2. But you aren't going to get that much because you didnt start with a whole mole of citric acid. Instead, the amount of CO2 will be the same fraction of 132 that 2.57 is of 192. Get it?

Answer 2

0.0133 mol C6H8O7 * ( 3 CO2 / 1 C6H8O7 ) = 0.041 mol CO2

0.041 mol * 44 g/mol = 1.77 gm