How to solve this limiting reactant problem

3Si + 4N2 --> 2Si3N4.
Determine the mass of silicon nitride produced when 5.00g of Si react with 2.50 grams of nitrogen gas.

It's a limiting reactant problem, I've tried it out, and I'm not 100% sure on my answers.
So if you could show me generalize steps too, that'd be great.

Ba (NO3)2 + Na2S04 --> BaSO4 + 2NaNO3
How many grams of barium sulfate can be produced from 235 g of Barium nitrate and 67.2 grams of Sodium Sulfate? Which is in excess? How many grams of the excess reactant remaining?

I could not figure this one out for the life in me. I've tried, but my numbers come out really weird.
Again, step by step would be so helpful too.
Thanks!

(a)

3 Si + 4 N2 ---> 2 Si3N4


To find the limiting reactant, we need to find the moles of each reactant:

moles = mass (g) / molecular weight (g/mol)


moles (Si) = 5.00 g / 28.09 g/mol = 0.178 moles
moles (N2) = 2.50 g / 28.01 g/mol = 0.089 moles


We must compare the two values using the molar ratios in the equation:

0.178 moles Si x (4 moles of N2 / 3 moles of Si) = 0.237 moles of N2

Comparing this value to the value of 0.089 moles of N2 that we calculated in the first part of this problem allows us to determine that we have fewer moles of N2 (when compared to Si). Therefore, the limiting reactant is N2.

Now, we multiply the amount of limiting reactant by the mole ratio of the product, Si3N4, to the reactant:

0.089 moles N2 x (2 moles Si3N4 / 4 moles N2) = 0.0445 moles of Si3N4

mass Si3N4 = moles Si3N4 x molecular weight Si3N4 = 0.0445 mol x 140.28 g/mol = 624.25 g



(b)

Ba (NO3)2 + Na2S04 --> BaSO4 + 2NaNO3


To find the limiting reactant, we need to find the moles of each reactant:

moles = mass (g) / molecular weight (g/mol)


moles [Ba(NO3)2] = 235 g / 261.34 g/mol = 0.899 moles
moles (Na2S04) = 67.2 g / 142.04 g/mol = 0.473 moles