calculate the volume of a 0.02M mno4- solution that is required to react completely with 10.0ml of 0.1M c2o4-2
Write the equation and balance it.
mols C2O4^2- = M x L = ?
Using the coefficients in the balanced equation, convert mols oxalate to mols permanganate. Then
M MnO4^- = mols/L. You know M and mols, solve for L.
molarity times liters? but where would i get liters from?
Read the problem.
mols oxalate = M x L
M = 0.1M
volume = 10.0mL = 0.010 L big as life right there in the problem.
To calculate the volume of the 0.02 M MnO4- solution required to react completely with 10.0 ml of 0.1 M C2O4-2, we need to use the stoichiometry of the balanced chemical equation for the reaction. The balanced chemical equation for the reaction between MnO4- and C2O4-2 is:
2MnO4- + 5C2O4-2 + 16H+ → 2Mn2+ + 10CO2 + 8H2O
From the balanced equation, we can see that the ratio of MnO4- to C2O4-2 is 2:5. Hence, for every 2 moles of MnO4- reacting, we need 5 moles of C2O4-2.
Step 1: Calculate the number of moles of C2O4-2 in the given volume:
Moles of C2O4-2 = Concentration of C2O4-2 × Volume of C2O4-2 solution
= 0.1 M × 0.01 L
= 0.001 mol
Step 2: Use the stoichiometric ratio to determine the number of moles of MnO4- required:
Moles of MnO4- = (5/2) × Moles of C2O4-2
= (5/2) × 0.001 mol
= 0.0025 mol
Step 3: Calculate the volume of the 0.02 M MnO4- solution required:
Volume of MnO4- = Moles of MnO4- / Concentration of MnO4-
= 0.0025 mol / 0.02 M
= 0.125 L
= 125 ml
Therefore, the volume of the 0.02 M MnO4- solution required to completely react with 10.0 ml of 0.1 M C2O4-2 is 125 ml.