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

3Si + 2N2 --> Si3N4 (as given)

n(Si) = m/MM = 38.25/28.085 = 1.3619 mol
n(N2) = 14.33/2*14.007 = 0.5115 mol

Therefore, N2 is limiting and Si is in excess
The molar ratio of 2N2:Si3N4 is 2:1
So, 0.0575 mol of silicon nitride is formed (dividing 0.5115 by 2)

m of silicon nitride= n*mm = 0.0575*140.283 = 8.06627... g
= 8.066g (4 significant figures)

(hopefully it is right, but double check in case i did something wrong) :)

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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

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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