# What is the limitting reactant? Ba(OH)2 * 8H20 (s) + 2NH4SCN (s) -> Ba(SCN)2 (s) + 10H2O (l) + 2NH3 (g)Each reactant is 10.0 g, which is the limitting reactant and how many moles of Ba(SCN)2 (s) is produced?

In order to identify the limiting reactant we have to identify which between the two reactant would produce the smaller amount of products.

We can look at it this way:

1.   Ba(OH)2 * 8H20  ------> moles Ba(SCN)2

2.   NH4SCN --------> moles Ba(SCN)2

There are two ways in which Ba(SCN)2 can be formed. The moles of Ba(SCN)2 will dictate which is the limiting reactant. The smaller Ba(SCN)2 formed means it is produced from the limiting reactant.

So solving them would produce the amount of moles.

1.

10g Ba(OH)2 * 8H20  x (1mole Ba(OH)2*8H2O/ 315.4650 g) x ( 1 mole Ba(SCN) / 1 mole Ba(OH)2*8H2O)

= 0.0317 moles Ba(SCN)2

2.

10 g NH4SCN x ( 1mole NH4SCN / 76.1215 g) x ( 1 mole Ba(SCN)2 / 2 moles NH4SCN)

= 0.0658 moles Ba(SCN)2

Comparing the two Ba(SCN)2 produced we can see that the moles of Ba(SCN)2 via  Ba(OH)2 * 8H20 is the smaller. Therefore Ba(OH)2 * 8H20 is the limiting reagent and the moles Ba(SCN)2 is 0.0317 moles.

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