Calculations+in+Chemical+reactions

= = toc = = = = = = =Calculating the Yield from a reaction =

Whenever you are given a chemical equation and the amount of reactant that you have, you can calculate the yield (or in other words the amount of the product that you get).

//Example: //

What mass of Iron do you expect to get when 4 kg of Iron oxide (Fe 2 O 3 ) reacts with Carbon monoxide (CO) according to the equation below:

Fe 2 O 3 + 3CO ---> 2Fe + 3CO 2

4 Kg of Fe is equal to 4000g

//Now work as follows: //

Find the number of moles of Iron Oxide that you have:

Moles of Fe 2 O 3 =

If you remember the previous lesson, we learnt that the number in front of each element in a reaction represents their molar ratio. So here for every 1 mol of Fe 2 O 3 you get 2 mol of Fe.

So the number of mol of Fe is twice that of Fe 2 O 3 = 25 x 2 = 50 mol

Thus, we get 50 mol of Fe from the reaction. We know the molar mass of Fe, so we can calculate the mass of Fe that we get= 50 x 56 = 2800g=2.8 Kg

According to our calculations you get 2.8Kg of Fe from 4Kg of Fe 2 O 3.

=Limiting and Excess Reagents =

Supposing you want to make a fruit salad and the rule is that for every plate of Salad that you make, you’ll need 2 cucumbers and one tomato. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Look at the above picture. You have 6 cucumbers and 5 tomatoes. So how many plates of salad can you make? Start off by taking two cucumbers and a tomato and see how many plates of salad you can make. As you see, we get three plates. All of the cucumbers have been used but we still have two tomatoes left. We can’t make any salad with the two because we will need more cucumbers. Thus the cucumbers are limiting the amount of Salad that we can get while the tomatoes are in excess.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Such kind of a situation happens when a chemical reaction takes place. One of the reactants is in excess and not all of it will react while the other one is limiting and all of it will react. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">So we define:

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">__ //**Excess Reagent**// __-We have more than necessary of this reagent, not all of it will react.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">__**// Limiting Reagent //**__-There is not enough of it present. All of the reagent will react

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Therefore, whenever you want calculate the yield in a reaction you need to use the reagent that is limiting and not the excess one. Look at the example below:

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Look at the following reaction involving the reaction of magnesium with hydrochloric acid. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Mg + 2HCl ---> MgCl 2 + H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Find the limiting and excess reagent? <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">First we need to find the number of moles of each reactant that we have:

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Moles of Mg = <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Moles of HCl =

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Now that we know the number of moles of each reactant we need to do some analysis as follows:

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If you wanted all the 0.05 mol of Mg to react, how many moles of HCl is required? <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Mol of HCl need = twice the mole of Mg(remember the mole ratio Mg:HCl = 1:2) = 2 x 0.05 = 0.1 mol

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Now check! According to the calculation if all the Mg is to react then we need //0.1 mol of HCl//. But, how many moles do we have? <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">We have //0.075 mol which is not enough//. So HCl is the limiting reagent while Mg is the excess reagent as some of it will not react.


 * //<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Alternatively, you could do the following: //**

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If you wanted all the 0.075 mol of HCl to react, how many moles of Mg do you need?

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Mol of Mg required = half the mole of HCl(remember the mole ratio Mg:HCl = 1:2) = ½ x 0.075 = 0.0375 mol

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Now Check! According to the calculation if all the HCl is to react then we need //0.0375 mol of Mg//. How many moles of Mg do we have? We have //0.1 mol of Mg//, which is more than enough. So Mg is the excess reagent while HCl is the limiting reagent.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">All the above working is done to show you how we can find the excess and limiting reagent. However there is shorter way of doing this as follows:



= **<span style="font-family: Tahoma,sans-serif;">Percentage yield: ** = <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Using the previous example. Find the amount of MgCl 2 that you get. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">We know that HCl is the limiting reagent. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Thus, number of mol of MgCl 2 = 0.075 x 0.5= 0.0375 mol <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Convert this into grams=0.0375 x 95.21 = 3.57 g

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">According to our calculations we expect to get 3.57 g of MgCl 2. This is called the theoretical yield since it has been calculated. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">But in real life, if you were to do the experiment practically, you would get a smaller amount than what you calculated. This is because maybe not all of the reactants reacted or other reasons.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Let’s say that you carried out the above reaction and got 2.5 g of MgCl 2. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Can you calculate the percentage yield?



<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">So we have a percentage yield of 70%. (We got 70% of what we expected!!)

= **<span style="font-family: Tahoma,sans-serif;">Avogadro’s hypothesis ** =

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Avogadro said that one mole of any gas, at equal temperature and pressure, occupies the same volume. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">What it means is that if you have one mole of nitrogen gas, one mole of oxygen gas, one mole of carbon dioxide gas and one mole of any other gas, at the same temperature and pressure conditions, then they will all occupy the same volume.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">In simpler terms, it means that the value in front of any gas in a chemical equations does not only show their molar ratio but also their volume ratio.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Look at the following equation: <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">2CO (g) + O 2 (g) ---> 2CO 2 (g) <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">From the equation we know that for 2 mol of CO we get 1 mol of O 2 and 2 mol of CO 2. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Likewise, we can say that for CO occupies twice the volume of O 2 and also the CO 2 produced occupies twice the volume of O 2.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">At different temperatures and pressures a gas would occupy different volumes. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">At the standard temperature (0 o C =273 K) and pressure (1 atm=101.3 kPa) One mole of any gas would occupy a volume of 22.4 dm 3. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">So one mole of oxygen, nitrogen, carbon dioxide and any other gas would occupy a volume of 22.4 dm 3 at the standard temperature and pressure (s.t.p)

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">You can use the following formula to find the number of moles of a gas when you are given the volume:



//<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Example: // <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">How many moles of oxygen are there in 5 dm 3 of oxygen at s.t.p (standard temperature and pressure) <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">We know that at s.t.p one mole of oxygen occupies a volume of 22.4 dm 3 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">So, number of moles of oxygen =

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">What volume of Nitrogen dioxide would you expect to get from 20 cm 3 of nitrogen monoxide.(all the volumes are measured at the same temperature and pressure) <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">2NO (g) + O 2 (g) ---> 2NO 2 (g) <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The volume ratio for NO and NO 2 are 1:1 so we will get 20 cm 3 of NO 2. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">How much O 2 do you need? <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The volume ratio of NO to O 2 is 2:1 so you will need 0.5 x 20= 10 cm 3 of O 2.
 * //<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Question: //**

= **<span style="font-family: Tahoma,sans-serif;">The ideal gas equation ** =

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">An ideal gas is a gas whose particles have a negligible volume and there are no attractive forces between the particles. Real gases tend to behave like an ideal gas at very low pressures and high temperatures. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The pressure, volume, temperature and the amount of and ideal gas are related by the following equation:

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">R is the called the idea gas constant and has a value of 8.31 J K -1 mol -1 //(This is given on page 2 of the data booklet)//

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">An ideal gas has pressure of 10 4 Pa, Temperature of 20 o C and a volume of 1 m 3. Find the amount of the gas present? <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">We know that **PV=nRT**

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Rearranging the formula gives= <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">So we can calculate the number of moles of the gas as follows: <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Note that the temperature has been converted into kelvins. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">You can convert from o C to Kelvin as follows:

**<span style="font-family: Tahoma,sans-serif;"> K = o C + 273 **

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Using the ideal gas equation, we have so many variables; pressure, temperature, volume and the number of moles. We can look at the relationship between any two of volume, pressure and temperature of an ideal gas. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">For a fixed mass of an ideal gas (that means the number of moles are constant):

//__<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">When temperature is constant: __// <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The pressure and volume are related as follows: <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">So we can say that: **P 1 V 1 = P 2 V 2 ** <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">P 1 and V 1 are the initial pressure and volume <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">P 2 and V 2 are the pressure and volume after a change //<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">A graph of P versus V looks like this //

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">

//__<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">When pressure is constant: __// <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The volume and temperature are related as follows: <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">So we can say that: <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">V 1 and T 1 are the initial temperature and pressure <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">V 2 and T 2 are the pressure and temperature after a change //<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">A graph of V versus T looks like this // <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> //__<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">When volume is constant: __// <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The pressure and temperature are related as follows: <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">So we can say that: <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">P 1 and T 1 are the initial temperature and pressure <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">P 2 and T 2 are the temperature and pressure after a change //<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">A graph of P versus T looks like this // <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">All of the above can be combined into one expression:

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">So if you are given the initial conditions of a gas and then there is a change of volume, temperature and/or pressure of the gas, then you can calculate the final temperature, volume or pressure of the gas. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">//Example:// The pressure on 600 cm 3 of a gas is increase from 100 KPa to 300 KPa at constant temperature. What will be the new volume of the gas?

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">V 1 = 600 cm 3 P 1 =100KPa V 2 = ? P 2 =300kPa <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">So we rearrange the formula and calculate V 2 as follows: <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">

**By the end of this lesson you should be able to:**
 * **Calculate the theoretical yield from a chemical equation**
 * **Find the limiting and excess reagents**
 * **Find percentage yield**
 * **Apply Avogadro's hypothesis to gases**
 * **know that the volume of one mole of gas at s.t.p is 22.4 dm 3 **
 * **Solve problems using the ideal gas equation**
 * **Be familiar with the graphs shown above**

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