Na2CO3 + 2HCL CO2 + H2O + 2NaCl
Reaction Ration is 1:2
Calculate how many moles of HCl are present in the average volume of HCl at the end-point.
Avg. volume is 0.0182L
mols = M x L = ? You have L but no M
or
mols Na2CO3 = grams/molar mass = ? Then mols HCl = mols Na2CO3 x 2 x L. Again, you have L HCl but no mols Na2CO3.
To find the number of moles of HCl present in the average volume, you need to use the reaction ratio and the volume of HCl.
Looking at the balanced chemical equation:
Na2CO3 + 2HCl → CO2 + H2O + 2NaCl
The reaction ratio between Na2CO3 and HCl is 1:2. This means that for every 1 mole of Na2CO3 reacted, 2 moles of HCl are required.
In this equation, we know the average volume of HCl is 0.0182 L. However, we need to convert this volume into moles of HCl. To do this, we'll use the ideal gas law equation:
PV = nRT
Where:
P = pressure (usually in atm, but we're assuming it's at STP, so it's 1 atm)
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L•atm/(mol•K))
T = temperature in Kelvin (273.15 K)
Since the pressure, temperature, and R are constant, we can rearrange the equation to solve for n:
n = PV / RT
Now plug in the values:
P = 1 atm
V = 0.0182 L
R = 0.0821 L•atm/(mol•K)
T = 273.15 K
n = (1 atm) * (0.0182 L) / (0.0821 L•atm/(mol•K) * 273.15 K)
n ≈ 0.00081 moles of HCl
So there are approximately 0.00081 moles of HCl present in the average volume of HCl at the end-point.