Assuming it is NaHCO3
In order to find the limiting reactant, we need to set up the reaction first:
10 NaHCO3 + 15 CH3COOH -> 10 NaCH3COO + 2 H2O + 20 CO2
Now we can see the molar ratio between NaHCO3 and acetic acid, which is 10:15, 10 moles of NaHCO2 for every 15 moles of acetic acid, now we have to find the number of moles of both compounds in order to find the limiting reactant
For NaHCO2, we will be using the given mass, 5.63 grams and the molar mass, 84g/mol
84g = 1 mol
5.63g = x moles
x = 0.067 moles of NaHCO3 in 5.63 grams, and according to the molar ratio, we will have:
10 NaHCO3 = 15 acetic acid
0.067 NaHCO3 = x acetic acid
x = 1.005 moles of acetic acid, this is how much we need of acetic acid in order for the reaction to occur
But we need to check if that is the value we have, or if we have more than that, we will do that by using the Molarity formula:
M = n/V
We have:
M = 0.83 M
V = 0.025 Liters
n = ?
0.83 = n/0.025
n = 0.021 moles
We have way less acetic acid than we actually need, which means, ACETIC ACID THE LIMITING REACTANT, and we have a lot of NaHCO3 in the reaction