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Calculating Percent Yield

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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,

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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)

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, you need twice as much iodine as tin, so even though you have more moles of iodine, it will be your limiting reagent:

Sn 0.00202 / 1 = 0.00202
I2 0.00374 / 2 = 0.00187 <-- limiting ...

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