Question

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?

Answer 1

• The limiting reagent is Sn(NO₂)₄

• The yield is 87.5%

Step-by-step explanation:

The reaction that takes place is

• 3Sn(NO₂)₄ + Pt₃N₄ → Sn₃N₄ + 3Pt(NO₂)₄

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

• Sn(NO₂)₄ ⇒ 302.71 g/mol

• Pt₃N₄ ⇒ 251.08 g/mol

• Sn₃N₄ ⇒ 412.13 g/mol

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

• 102 g Sn(NO₂)₄ ÷ 302.71 g/mol = 0.337 mol Sn(NO₂)₄

• 80.2 g Pt₃N₄ ÷ 251.08 g/mol = 0.319 mol 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:

• 40.2 g Sn₃N₄ ÷ 412.13 g/mol = 0.098 mol Sn₃N₄

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

• 0.337 mol Sn(NO₂)₄ *
  `(1molSn_(3)N_(4))/(3molSn(NO_(2))_(4))` = 0.112 mol Sn₃N₄

So the percent theoretical yield is:

• 0.098 / 0.112 * 100% = 87.5%