Calculate the mass of Na2C2O4 needed to react with 30mL of 0.050 M KMnO4.
The net ionic equation is:
2MnO4- + 5C2O4-2 + 16H+ —> 2Mn+2 + 10CO2 + 8H2O
I don't know where to start or how to end up with sodium oxalate.
Thanks!
Step1. Calculate the mols KMnO4 used. mols KMnO4 = M x L = ?
Step2. Using the coefficients in the balanced equation, convert mols KMnO4 used to mols Na2C2O4 needed.
Step3. Then mass Na2C2O4 = mols Na2C2O4 x molar mass Na2C2O4.
Well, let's start by figuring out the stoichiometry of the reaction. According to the net ionic equation you provided, two moles of KMnO4 react with five moles of C2O4-2. So we have a 2:5 ratio of KMnO4 to C2O4-2.
Now, let's calculate the moles of KMnO4 you have. You have 30 mL of a 0.050 M solution of KMnO4. To get moles, we can use the formula: moles = concentration (M) x volume (L).
Moles of KMnO4 = 0.050 M x (30/1000) L = 0.0015 mol
Now, using the 2:5 ratio, we can find the moles of C2O4-2 needed to react with the moles of KMnO4 you have.
Moles of C2O4-2 = (0.0015 mol KMnO4) x (5 mol C2O4-2 / 2 mol KMnO4) = 0.00375 mol C2O4-2
Finally, to find the mass of Na2C2O4 needed, we need to use the molar mass of Na2C2O4, which is approximately 134.00 g/mol.
Mass of Na2C2O4 = (0.00375 mol C2O4-2) x (134.00 g/mol) = 0.503 g
Therefore, you would need approximately 0.503 grams of Na2C2O4 to react with 30 mL of 0.050 M KMnO4.
But remember, this is just a theoretical calculation. In practice, you may need to use more or less of the Na2C2O4 depending on the reaction conditions, so be sure to double-check with a reliable source or follow a lab procedure. And of course, always wear your safety goggles when working with chemicals. Hope this helps, and good luck with your experiment!
To calculate the mass of Na2C2O4 needed to react with 30 mL of 0.050 M KMnO4, you need to follow a series of steps:
1. Calculate the moles of KMnO4 used:
Moles of KMnO4 = Volume (L) x Concentration (M)
Note: Convert the volume from mL to L (divide by 1000).
Given: Volume = 30 mL = 0.030 L
Concentration = 0.050 M
Moles of KMnO4 = 0.030 L x 0.050 M = 0.0015 moles of KMnO4
2. Use the stoichiometry of the balanced equation to determine the moles of Na2C2O4 needed. From the net ionic equation, we know that 5 moles of C2O4-2 react with 2 moles of KMnO4. Therefore:
Moles of C2O4 = (Moles of KMnO4) x (5 moles of C2O4-2 / 2 moles of KMnO4)
= 0.0015 moles of KMnO4 x (5/2)
= 0.00375 moles of C2O4-2
3. Convert the moles of C2O4-2 to the mass of Na2C2O4 using its molar mass:
Molar mass of Na2C2O4 = 2(22.99 g/mol for Na) + 2(12.01 g/mol for C) + 4(16.00 g/mol for O)
= 2(22.99 g/mol) + 2(12.01 g/mol) + 4(16.00 g/mol)
= 46.00 g/mol + 24.02 g/mol + 64.00 g/mol
= 134.02 g/mol
Mass of Na2C2O4 = Moles of C2O4-2 x Molar mass of Na2C2O4
= 0.00375 moles of C2O4-2 x 134.02 g/mol
= 0.5027 g
Therefore, the mass of Na2C2O4 needed to react with 30 mL of 0.050 M KMnO4 is approximately 0.5027 grams.