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What are the limiting and excess reactant when 600g of Si reacts with 500g of N2

What are the limiting and excess reactant when 600g of Si reacts with 500g of N2-example-1

1 Answer

4 votes

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

-----

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

---------------

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

answered
User Erb
by
7.6k points
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