calculating mass of an excess reagent?

Question

calculate the mass of excess reagent remaining at the end of the reaction in which 90.0g of SO2 are mixed with 100.0g of O2.
SO2+O2--> SO3

can someone help me with this, i am kind of stuck, also can you show me how you got your answer, so i can see where i went wrong.

Answer 1

first you need to balance the equation:

2SO2+O2=2SO3

This gives you the mole ratio of the components. First, figure out which one is the limiting reactant. I can't think of a way to show this calculation on here, but basically you start with the grams of the reactant, convert it to moles using the molar mass, use the mole ratio to switch to the product, and then convert back to grams using the molar mass of the product.
You have to do this twice using the other reactant the second time.
Once you know which one is limiting you take the mass of it, convert it to moles, then take the mole ratio of the reactants and switch to moles of the other reactant, then switch back to grams. You subtract your final mass from how much you started with and this will give you the excess amount of the reactants in grams. email me if you need more help.

p.s. The first answerer was wrong, that isn't how mole ratios work.

Answer 2

Remember, you have to do your comparisons on a mole basis. So convert both SO2 and O2 masses to moles (about 1.2 and 3 respectively). Since 1 mole of SO2 requires 1 mole of O2, the O2 should be in excess. The moles in excess are the moles you started with less those involved in the reaction. Then you convert back to mass units.