Limiting Reactants/Chemistry Problem

Determine the weight of H2SeO4 that will be formed by reaction of 100.0 g of H2O, 150.0 g of Se and 100.0 g of O2 via the reaction 2H2O + 2Se + 3 O2  2 H2SeO4

For H2SeO4 weight I got 245.022g. Is that correct?

You are in the right ball park, which indicates to me that you've determined the correct limiting reagent, but without seeing your work I can't tell if you have just made a typo or if you have made some other mistake.

Determine the limiting reagent... Do this by working out how much H2SeO4 would form from all of each reagent. The one that produces the least is the limiting reagent.

moles H2O = mass / molar mass = 100.0 g / 18.016 g/mol = 5.551 mol
2 moles H2O reacts to give 2 moles H2SeO4
Therefore 5.551 mol H2O can produce 5.551 mol H2SeO4

moles Se = 150.0 g / 78.96 g/mol = 1.900 mol
2 moles Se reacts to give 2 moles H2SeO4
Therefore 1.900 mole Se can give 1.900 moles H2SeO4

moles O2 = 100.0 g / 32.00 g/mol = 3.125 mol
3 moles O2 react to give 2 moles H2SeO4
Therefore moles H2SeO4 = 2/3 x moles O2 = 2/3 x 3.125 mol = 2.083 moles

The Se provided produces the least H2SeO4, therefore Se is the limiting reagent.

Maximum amount of H2SeO4 possible is from complete reaction of all the limiting reagent. Se is limiting, and we worked out above that 1.900 moles of Se can produce 1.900 moles H2SeO4.

mass H2SeO4 = moles x molar mass = 1.900 mol x 144.976 g/mol = 275.4544 g
= 275.5 g ( 4 sig figs)