1. Stoichiometry

2. Reading Equations using Moles Ca + H 2 SO 4 CaSO 4 + H 2 The above reaction reads: 1 amount of Ca reacts with 1 amount of H 2 SO 4 to give 1 amount of CaSO 4 and 1 amount of H 2 gas. (we don’t usually put in the ones – but if we did it would look like this) The amount we use is the mole Now the reaction reads: 1 mole of Ca reacts with 1 mole of H 2 SO 4 to give 1 mole of CaSO 4 and 1 mole of H 2 gas. 1111

3. Reading Equations using Moles Ca + H 2 SO 4 CaSO 4 + H 2 Questions: 1.If we have 1 mole of Ca how many moles of H 2 SO 4 are needed to fully react the Ca? 2.If we had 0.5 mole of Ca how many moles of H 2 SO 4 would be required? 3.If 0.5 mole of Ca reacted with 0.5 mole of H 2 SO 4 how many moles of CaSO 4 and how many moles of H 2 gas are formed? Ans: 1mole Ans: 0.5 mole Ans : 0.5 mole of H 2 gas,0.5 mole of CaSO 4

4. Reading Equations using Moles Ca + H 2 SO 4 CaSO 4 + H 2 Questions: If 0.5 mole of Ca reacted with 0.5 mole of H 2 SO 4 how many moles of CaSO 4 and how many moles of H 2 gas are formed? Ans : 0.5 mole of H 2 gas,0.5 mole of CaSO 4 Given the Ar (atomic mass) values H = 1 gmol -1 Ca = 40 gmol -1 S = 32 gmol -1 O =16 gmol -1 Find the molar masses (Mr) of H 2 SO 4, CaSO 4 and H 2 Then use these and the mole (n) formula to work out the mass of H 2 gas and the mass of CaSO 4 formed (Hint you will need to rearrange the formula)

5. Reading Equations using Moles Fe 2 O 3 + 3CO 2Fe + 3CO 2 Q1. If 200g of Fe 2 O 3 is reacted in the above equation what mass of Fe is produced? (Mr Fe 2 O 3 = 160 gmol -1, Mr Fe = 55.85 gmol -1 ) From the equation we see that for every mole of Fe 2 O 3 that reacts 2 moles of Fe are formed) This means that if we can find the number of moles of Fe 2 O 3 we can then work out the number of moles of Fe and then the mass of Fe Using the mole ratio from the equation how many moles of Fe are formed? (unknown/known) You got it 2.5 moles of Fe is formed –now work out the mass of Fe

6. Reaction ratio Mg + 2 HCl MgCl 2 + H 2 Work out how many moles of HCl is used to completely react 60grams of Mg metal in the following reaction? Draw the grid 2 mass Mg ? 60.0 grams = 2.5 moles 24.0 gmol -1 Mr HCl 36.5 gmol -1 1 Therefore (unknown/known) 2mol / 1mol x moles of Mg = moles of HCl Therefore 2 x 2.5 moles = 5 moles of HCl 5.00 moles 183 grams

7. What was the mass of HCl used? m = 5 x Mr = 5 mol x 36.5 gmol -1 = 182.5 g = 183 g (3sf)