In a certain experiment 1.00 g of sodium bicarbonate and 1.00 g of citric acid are allowed to react. (a) which is the limiting reactant? (b) how many grams of carbon dioxide form? (c) how many

The complete balanced chemical reaction for this is:

C6H8O7 + 3 NaHCO3 - - - > 3 H2O + 3 CO2 + Na3C6H5O7

First calculate for the number of moles of each reactant.

moles C6H8O7 = 1 g / (192.124 g / mol) = 5.2 * 10^-3 mol

moles NaHCO3 = 1 g / (84.01 g / mol) = 11.9 * 10^-3 mol

A. The ratio of the reactant from the chemical reaction is 3NaHCO3:1C6H8O7, while the given chemicals are in the ratio of:

11.9 * 10^-3NaHCO3: 5.2 * 10^-3 C6H8O7 = 2.29NaHCO3:1C6H8O7

Therefore this means that there is less amount of NaHCO3 supplied than what is required therefore the limiting reactant is:

NaHCO3

B. We calculate based on the limiting reactant.

mass CO2 = 11.9 * 10^-3 mol NaHCO3 (3 mol CO2/1 mol NaHCO3) (44.01 g/mol)

mass CO2 = 1.57 g

C. I believe what is asked here is the amount of excess reactant which remains. The excess reactant is C6H8O7.

mass C6H8O7 left = [5.2 * 10^-3 mol - (11.9 * 10^-3 mol NaHCO3 (1 mol C6H8O7 / 3 mol NaHCO3)) ] * (192.124 g / mol)

mass C6H8O7 left = 0.237 g