For the reaction shown here, 2.3 mola is mixed with 1.6 molb and 3.1 molc. what is the limiting reactant? 3a+b+2c→3d

We know that the reactant ratios

a : b : c

are

3 : 1 : 2

So we will check the amount of the other two reactants required by taking one as the base.

With a as the base, 

2.3 : 0.77 : 1.53

The amounts of b and c required for 2.3 moles of a are present

With b as the base,

4.8 : 1.6 : 3.2

Both a and c are present in amounts less than that required for 1.6 moles of b.

With c as the base, 

4.65 : 1.55 : 3.1

Only a is limiting in this case

From the above considerations, we see that a is always in quantities less than that which would be required, so a is the limiting reactant.

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