38.25 grams of silicon is combined with 14.33 grams of nitrogen gas. how many grams of silicon nitride can be formed if nitrogen is the limiting reactant?
3si + 2n2 yields si3n4

Since it is stated that nitrogen is the limiting reactant (sometimes called limiting reagent), you only have to worry about the amount of nitrogen since we know that there is excess silicon.
Since this is a stoichiometry question, the first thing to do is turn the mass of nitrogen gas into moles by dividing the mass of nitrogen gas by its molar mass.  (14.33g)/(28g/mol)=0.5118mol nitrogen gas.  Then you have to convert moles of nitrogen gas to moles of silicon nitride by using the fact that 2 moles of nitrogen gas turn into 1 mole of silicon nitride (i got that ratio from the chemical equation given).  (0.511mol N₂)x(1mol Si₃N₄/2mol N₂)=0.2555mol Si₃N₄.  Know you just need to find the mass of silicon nitride by multiplying the number of moles of silicon nitride by its molar mass.  (0.2555mol)x(140g/mol)=35.85g
Therefore the amount of silicon nitride produced from this reaction 35.85g.

I hope this helps

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