1 / 14

Chemical quantities in reactions

Chemical quantities in reactions. What and why. If we know the chemical equation for a product: We can work out how much we need We can work out how much we will get We can work out the percentage efficiency of processes Why is all this so important in the real world?. Starter questions.

bianca
Download Presentation

Chemical quantities in reactions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chemical quantities in reactions

  2. What and why If we know the chemical equation for a product: We can work out how much we need We can work out how much we will get We can work out the percentage efficiency of processes Why is all this so important in the real world?

  3. Starter questions 1. What is the mass of 0.5moles of water? 2. What are the products of this equation? Ca + 2HCl = 3. Balance this equation Na + Cu3(PO4)2 = Na3PO4 + Cu 4. Complete and balance this double substitution reaction. Ag2SO4 + Pb(NO3)4 =

  4. Calculations in equations Calculate the mass of reactants and products when 1mole of C burns in oxygen. (What do we know about the mass of reactants and products from the start?) C + O2 = CO2 Is this balanced? Molar masses C = 12g, O2 = 32g 12 + 32 = 44g = 1 mole of CO2 Amount of reagentsis 44g. Amount of products is 44g

  5. Do these 2 moles Iron + 3 moles Sulphur 2 mole Sodium hydroxide + 2 mole Hydrochloric acid

  6. Moles of a reactant needed. How many moles of sulphur are needed to react with 1.42 moles of iron (iii)? Write equation 2Fe + 3S = Fe2S3 Ratio of Iron to Sulphur is 2:3 3/2 x amount of iron = amount of sulphur 1.5 x 1.42 = 2.13moles = amount of sulphur 2.13 x 32g = 68.16g = mass of sulphur needed.

  7. Limiting reactants If you don’t have enough of one reactant it will limit how much product is produced. Making hydrogen chloride Equation H2 + Cl2 = 2HCl So if we have one mole of each reactant we can get 2 moles of the product But if we have 2 moles of hydrogen and 5 moles of chlorine how much can we get? Which is limiting How much product can we make per mole? So how much product can we make in total?

  8. Limiting masses Ammonia and fluorine react to produce dinitrogentetrafluoride and hydrogen fluoride. NH3 + F2 = N2F4 + HF (balance this!) If 5g of NH3 and 20g of F2 react how many g of HF are produced Steps:

  9. Steps 5 g NH3 Molar Mass, Moles NH3 mole-mole factor, moles HF, Molar mass, grams HF 20g F2 Molar mass, Moles F2 mole-mole factor, moles of moles HF, molar mass, grams HF Molar mass of NH3 = 17g. We have 5g so we have 0.294 moles. 2 moles of NH3 make 6 moles of HF so 0.294 moles will make 3 x as much HF i.e. 0.882 moles

  10. next Molar mass of F2 = 38g 20g of F2 = 20/38 = 0.526moles of F2 5 moles of F2 make 6 moles of HF So 0.526 moles of F2 will make 6/5 x 0.526 moles of HF, i.e. 0.632 moles of HF This is fewer moles of HF than can be made from the ammonia so the fluorine is the limiting factor. (I will post exercises on blackboard)

  11. And finallyPercentage yields In reality no chemical reactions work perfectly. For this reason it is important to work out the percentage yield by dividing the actual yield by the theoretical yield you would get if it were 100% efficient. Percentage yield = actual yield / theoretical yield x 100.

  12. Example Equation LiOH + CO2 = LiHCO3 What is the % yield if 50g LIOH gives 72.8g LiHCO3 Equation is balanced. 1 mole LiOH gives 1 mole LiHCO3 50g / molar mass of LiOH = 50 / 24 = 2.08moles. As mole factor is 1:1 reactant to product then it should give 2.08 moles of LiHCO3

  13. Moving on 1 mole of LiHCO3 has a mass of 68g 2.08 moles of LiHCO3 has a mass of 141.67g % yield = actual yield /theoretical yield = 72.8/141.67 = 51.4%

  14. There are plenty of activitiestokeepusbusy!

More Related