# Calculating Percent Yield

1.Sn(IV)4?
-my experimental value was 0.194g and my melting point: 147 C
- my balanced equation Sn + 2I2 -------> SnI4
- we used .24g of Sn, 0.95g I2
(I got 1.25 g as my theoretical value didn't look right so I need to double check)

2.Sn(II)I2?
-my experimental value was 0.314g and melting point 160 C
- balanced equation Sn+2 (aq) + 2I- -----> SnI2(s)
-we used 0.32g Sn, 0.40g Iodine
( I got 0.324 for this one so i dont know if its right)

3.[(NH4)2]SnCl6
-my experimental value was 1.20g and 160 C is melting point
-Sn(IV)Cl4 +2NH4Cl--------> (Sn(IV)Cl6)^-2 + (NH4)2
-we used 1ml anhydrous SnCl4 and 0.5 ml of distilled water adeded to SnCl4 , 0.14g NH4Cl dissolved in 1ml of water,

© SolutionLibrary Inc. solutionlibary.com 9836dcf9d7 https://solutionlibrary.com/chemistry/physical-chemistry/calculating-percent-yield-7e8a

#### Solution Preview

...gent

Now you use the limiting reagent to predict moles:

Now you can convert moles to grams to find expected yield:

Finally you divide experimental by expected: 0.24 g / 1.17 g = 21%

2. Sn(II)I2?
-my experimental value was 0.314g and melting point 160 C
- balanced equation Sn+2 (aq) + 2I- -----> SnI2(s)
-we used 0.32g Sn, 0.40g Iodine
( I got 0.324 for this one so i dont know if its right)

To find the percent yield, first you need to find the moles of each reactant:

Now that you know the moles, you need to know the limiting reagent. According to your balanced reaction, the ratio of iodine (I2  2 I-) is even with tin, so you base it completely on the moles above.

Sn 0.00270 / 1 = ...