What mass of sodium oxalate na2c2o4 is needed to prepare 0.250 l of a 0.100 m solution?

Molarity = number of moles of solute / liters of solution
number of moles = molarity * liters of solution
number of moles of Na2C2O4 = 0.1 * 0.25 = 0.025 moles

Now, from the periodic table:
mass of Na = 22.9 grams
mass of C = 12 grams
mass of O = 16 grams
molar mass of Na2C2O4 = 2(22.9) + 2(12) + 4(16) = 133.8 grams
Therefore, one mole is equal to 133.8 grams. To know the mass of 0.025 moles, all you have to do is cross multiplication as follows:
mass = (0.025*133.8) / 1 = 3.345 grams

4 C2H5 + 13 O2 -> 8 CO2 + 10 H2O

Explanation:

What I do is I make a chart of how many elements the product side and reactant sides have:

Reactants-

2 C

5 H

2 O

Products-

1 C

3 O

2 H

It's easier to see how many coefficients are needed for the reaction to be balanced then. Hope this helped.