Step 1
the reaction must be completed and balanced as follows:
3 Si + 2 N2 => Si3N4
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Step 2
Information provided:
600 g of Si
500 g of N2
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Information needed: (use your periodic table please)
1 mole of Si = 28.0 g
1 mole of N = 14.0 g => 1 mole of N2 = 28.0 g
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Step 3
By stoichiometry,
3 Si + 2 N2 => Si3N4
3 x 28.0 g Si --------- 2 x 28 g N2
600 g Si --------- X
X = 600 g Si x 2 x 28 g N2/3 x 28.0 g Si
X = 400 g N2
For 600 g of Si, 400 g of N2 is needed, but there is only 500 of N2 so the excess is N2 and the limiting reactant is Si.
Answer:
The limiting reactant = Si
The excess reactant = N2
