Question

Write equations for the formation of SnI4 and (Ph3PO)2SnI4. Calculate your percentage yields.

- I am having trouble balancing this equation/calculating the yield

Why are stannic halides rendered more inert by complexation?

- Is this due to steric hinderance?

Answer 1

Solution:

Answer to Question No. 1:

The balanced chemical equation for reaction for the reaction of tin (Sn) and iodine (I₂) to form SnI₄ can be written as

SnI₄(Stannic iodide)

Sn + 2 I₂ → SnI₄

The balanced chemical equation for reaction for the reaction of SnI₄ with Ph3PO to form (Ph₃PO)₂SnI₄ can be written as

SnI₄ + 2 (Ph₃PO) = (Ph₃PO)₂SnI₄

Calculation of percentage yield: Typically in chemical reactions between two reagents, both are not used completely. In general, one may be used completely while some amount of the other reagent(s) may remain after the reaction has occurred. Those that remain are said to react in excess. The reactant that is used completely is the limiting reagent. The limiting reagent determines the amount of product(s) at the end of the reaction as well as how much remains of the reactant(s) that occurred in excess.

In order to determine which reactant is the limiting reagent, take each reactant separately and assume that it is the limiting reagent. The reactant that produces the least amount of product must be the limiting reagent.

To calculate the percentage yield we need to know the amount of reactants used. In the given problem the amount of reactants used is not provided so I am using a hypothetical value for the calculation.

Tin (Sn) and iodine (I2) react to produce tin iodide (SnI4). I am assuming that we have taken 1.00 g of tin and 2.5 g of I₂ at the beginning of the reaction. The balanced chemical equation is Sn + 2 I₂ → SnI₄

In this case, we do not know which reactant is the limiting reagent. There are many ways to determine the limiting reagent. We will show one method that involves assuming one reactant is the limiting reagent and determine whether or not there is a sufficient amount of the other reagent(s) for the assumed limiting reagent to react completely. So to start let's look at Sn and assume for now that it is the limiting reagent. How much iodine is needed to completely consume 1.00 g of Sn?

Mass of iodine needed = 1.00 g Sn × (1 mol Sn/118.710 g Sn) × (2 mol I₂/1 mol Sn)×(253.810 g i₂/1 mol I₂) = 4.27 g I₂

This means that in order for 1.00 g of Sn to be used completely, 4.27 g of iodine is needed. But we only have 2.5 g of I₂ (i.e., not enough I₂ to completely react with all of the tin). This means iodine is the limiting reagent.

Determine the amount of SnI₄ that may be produced: We must use iodine for this calculation since it is the limiting reagent.

Mass of SnI₄ produced:

= 2.5 gI₂ × ( 1 mol I₂/253.810 g I₂)×1 mol SnI₄ /2 mol I₂ ) ×(626.330 SnI₄/ 1 mol SnI₄) = 3.08 g SnCl₄

Therefore the theoretical yield is 3.08 g SnCl4.

Percent yield = (Actual yield/Theoretical yield)x 100

Actual yield can be obtained from experimental data.

Actual yield is not provided so it is not possible to calculate the percentage yield.

Note: Atomic mass of Iodine = 126.990 : Atomic mass of Sn = 118.71,Molar mass of SnI4 = 626.330

Molar mass of Ph₃PO = 278.285, Molar mass of (Ph₃PO)₂SnI₄ = 1182.898

Similarly, the percentage yield of (Ph₃PO)₂SnI₄ can be calculated by using the amount of reagents used.

Ans to Question no 2:

Stannic halides becomes more inert on complexation because the complexing "agent" changes the level of ionisation of the stannic halide. Moreover if triphynylphosphine ligand is used which is a poor leaving group, and possibly, it being a large ligand, there is considerable steric hindrance slowing attack by incoming species (ligands).

Note: A sample problem for the understanding of percentage yield is shown below.

Sample problem Q-What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2? Step 1: write the balanced chemical equation 2 Step 2: determine actual and theoretical yield Actual is given, theoretical is calculated: 2 moH 18.02 g H29 143 g 18 02 143 9 Step 3: Calculate % yield % yield = 138 g H2Ox 100%-96.7% theoretical x 100%--138g 143 g H2O