3NaHCO3(s) + C6H8O7(s) = 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq)
Who is the limiting reactant if you had 100g each of sodium bicarbonate and citric acid?
To determine the limiting reactant, we need to compare the number of moles of each reactant and see which one is used up completely or "limits" the reaction.
First, let's find the molar mass of each compound:
- Sodium bicarbonate (NaHCO3):
- Na = 22.99 g/mol
- H = 1.01 g/mol
- C = 12.01 g/mol
- O = 16.00 g/mol
- Total molar mass = 22.99 + 1.01 + 12.01 + (3 * 16.00) = 84.00 g/mol
- Citric acid (C6H8O7):
- C = 12.01 g/mol
- H = 1.01 g/mol
- O = 16.00 g/mol
- Total molar mass = (6 * 12.01) + (8 * 1.01) + (7 * 16.00) = 192.14 g/mol
Next, let's calculate the number of moles for each compound using the given mass:
- Moles of sodium bicarbonate (NaHCO3) = (mass of sodium bicarbonate / molar mass of NaHCO3)
- Moles of citric acid (C6H8O7) = (mass of citric acid / molar mass of C6H8O7)
Moles of sodium bicarbonate = (100 g / 84.00 g/mol) = 1.19 mol
Moles of citric acid = (100 g / 192.14 g/mol) = 0.52 mol
Since the reaction equation shows a 1:1 molar ratio between sodium bicarbonate and citric acid, we can conclude that the limiting reactant is the one with the lesser number of moles. In this case, the citric acid is the limiting reactant because it has 0.52 mol compared to the 1.19 mol of sodium bicarbonate.
Therefore, if you had 100g each of sodium bicarbonate and citric acid, the citric acid would be the limiting reactant.