If 102 grams of Sn(NO2)4 is reacted with 80.2 grams of Pt3N4, and yields 40.2 grams of Sn3N4. What is the limiting reagent in the reaction and what is the percent theoretical yield?

Explanation:

The reaction that takes place is

Now we write down the molar mass of Sn(NO₂)₄, Pt₃N₄ and Sn₃N₄:

In order to determine the limiting reagent, we calculate the moles of Sn(NO₂)₄ and Pt₃N₄:

From the balanced reaction we know that 1 mole of Pt₃N₄ would react completely with 3 moles of Sn(NO₂)₄, so 0.319 moles of Pt₃N₄ would require (0.319*3) 0.957 moles of Sn(NO₂)₄ to completely react.

There are not so many Sn(NO₂)₄ moles so Sn(NO₂)₄ is the limiting reagent.

To calculate the yield, first we calculate the moles of Sn₃N₄ produced:

And we compare it with the amount that the limiting reagent (Sn(NO₂)₄) could produce:

So the percent theoretical yield is:

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