you are given the equation
S2O3 + 2H --> S + SO2 + H2O
you mix 25 ml of HCl with a concentration of 2 mol/L
with 5 mL of Na2S2O3 with a
concentration of 0.3 mol/L.
what is the concentration of
S2O3 and H?
You REALLY should put charges on these equations. I spent 15 minutes trying to figure out what S2O3 was.
(S2O3^-2) = 0.3 x (5/20) =
(H^+) = 0.2 x (25/30) =
To determine the concentration of S2O3 and H, we can use the concept of stoichiometry and the balanced chemical equation given.
First, let's determine the moles of HCl used. We are given that 25 mL of HCl with a concentration of 2 mol/L is used. We can calculate the moles of HCl by using the formula:
moles = concentration * volume (in liters)
moles of HCl = 2 mol/L * (25 mL / 1000 mL/L) = 0.05 mol
Next, let's determine the moles of Na2S2O3 used. We are given that 5 mL of Na2S2O3 with a concentration of 0.3 mol/L is used. We can calculate the moles of Na2S2O3 by using the same formula:
moles of Na2S2O3 = 0.3 mol/L * (5 mL / 1000 mL/L) = 0.0015 mol
Looking at the stoichiometric coefficients in the balanced equation, we can see that 1 mole of H reacts with 1 mole of S2O3. Therefore, the moles of S2O3 present will be equal to the moles of Na2S2O3 used, which is 0.0015 mol.
Since there are no coefficients in the balanced equation indicating the relationship between H and HCl, we cannot directly determine the moles of H produced. However, we can use the balanced equation to determine the relationship between the moles of S2O3 and H.
From the balanced equation:
1 mol S2O3 produces 1 mol H
Therefore, the moles of H produced will also be 0.0015 mol.
To find the concentration of S2O3 and H, we need to determine the total volume of the solution after mixing the two substances.
The volume of HCl used is 25 mL, and the volume of Na2S2O3 used is 5 mL. Thus, the total volume of the solution will be:
25 mL + 5 mL = 30 mL = 0.03 L
Finally, we can calculate the concentration of S2O3 and H using the formula:
concentration = moles / volume
concentration of S2O3 = 0.0015 mol / 0.03 L = 0.05 mol/L
concentration of H = 0.0015 mol / 0.03 L = 0.05 mol/L
Therefore, the concentration of S2O3 and H is both 0.05 mol/L.