Calculating reacting masses Learning Objectives Aiming for 4

Calculating reacting masses

Learning Objectives • Aiming for 4: Calculate the masses of substances shown in a balanced symbol equation. • Aiming for 6: Calculate the masses of reactants and products from the balanced symbol equation and the mass of a given reactant or product. • Aiming for 8: Use given masses of reactants and products to create a balanced symbol equation for the reaction.

Reacting Masses • The masses of reactants and products can be calculated from balanced symbol equations. • Chemical equations can be interpreted in terms of moles. For example: Mg + 2 HCl Mg. Cl 2 + H 2 • This shows that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of magnesium chloride and 1 mole of hydrogen gas.

Q: Why might we want to know the mass of products of a reaction? Think about industry.

It’s important to be able to work out the mass of products of a reaction from the reactants because in industry, companies need to know how much product they’re making, and so how much profit they’re making and how much should be expected. It also helps measure percentage yield. You need to be able to calculate the mass of products in a reaction, so listen carefully to the next section.

Method for mass of a product 1. Write out the balanced equation for the reaction. 2. Find the Mr of the reactant and the product that you’re interested in. 3. Apply the rule: Divide to get one, then multiply to get all: § Divide both relative formula masses by the Mr of the reactant — this tells you how much product would be formed from 1 g of the reactant. § Multiply this by the amount of reactant given in the question to find out how much product would be formed from that much reactant. Copy into your book!

For example… • Q: What mass of magnesium oxide is produced when 60 g of magnesium is burned in air?

1. Write out the balanced symbol equation.

2. The reactant you’ve got is 2 Mg. The product you want is 2 Mg. O.

3. You need to know how much magnesium oxide can be made from 1 g of magnesium. So start by dividing both relative formula masses by the Mr of the reactant (that’s the magnesium).

4. The question asks you to find the mass of magnesium oxide produced when 60 g of magnesium is burned in air. So now you need to multiply both sides by 60. So there you have it — 60 g of Mg will produce 100 g of Mg. O.

Click to see the periodic table. Practice Question Q: What mass of calcium chloride (Ca. Cl 2) is produced when 24 g of calcium hydroxide (Ca(OH)2) reacts with an excess of hydrochloric acid (HCl)? A: 24 g of Ca(OH)2 will produce 36 g of Ca. Cl 2

Calculating reactant mass • You can use the same basic method to find how much reactant you’d need to use to make a certain mass of product. • The only difference is in Step 3 — this time you divide both relative formula masses by the Mrof the product. • This tells you how much reactant you would need to form 1 g of product. Then you just multiply everything by the mass of product given in the question.

Click to see the periodic table. For example… Q: How much zinc carbonate (Zn. CO 3) would need to decompose to form 24. 2 g of zinc oxide (Zn. O)? Try and work it out before we go through it together.

1. Write out the balanced symbol equation.

2. The reactant you want is Zn. CO 3. The product you’ve got is Zn. O.

3. This time, you want to know how much zinc carbonate you would need to produce 1 g of zinc oxide. So start by dividing both relative formula masses by the Mr of the product (zinc oxide).

4. Now all you have to do is multiply both sides by 24. 2 to find how much zinc carbonate you would need to make 24. 2 g of zinc oxide. So, 37. 3 g of Zn. CO 3 would need to decompose to form 24. 2 g of Zn. O.

Questions • Answer the questions on the sheet in your book.

Answers – Q 1 Mr of 2 KBr = 2 × (39 + 80) = 238 Mr of 2 KCl = 2 × (39 + 35. 5) = 149 238 g of KBr reacts to give 149 g of KCl (÷ 238) 1 g of KBr reacts to give 0. 626 g of KCl (× 36. 2) 36. 2 g of KBr reacts to give 22. 7 g of KCl

Answers – Q 2 Mr of 6 HCl = 6 × (1 + 35. 5) = 219 Mr of 2 Al. Cl 3 = 2 × (27 + (3 × 35. 5)) = 267 219 g of HCl reacts to give 267 g of Al. Cl 3 (÷ 219) 1 g of HCl reacts to give 1. 22 g of Al. Cl 3 (× 15. 4) 15. 4 g of HCl reacts to give 18. 8 g of Al. Cl 3

Answers – Q 3 Mr of Ca. CO 3 = 40 + 12 + (3 × 16) = 100 Mr of Ca. SO 4 = 40 + 32 + (4 × 16) = 136 100 g of Ca. CO 3 reacts to give 136 g of Ca. SO 4 (÷ 100) 1 g of Ca. CO 3 reacts to give 1. 36 g of Ca. SO 4 (× 28. 5) 28. 5 g of Ca. CO 3 reacts to give 38. 8 g Ca. SO 4

Answers – Q 4 Mr of KOH = 39 + 16 + 1 = 56 Mr of KNO 3 = 39 + 14 + (3 × 16) = 101 56 g of KOH reacts to give 101 g of KNO 3 (÷ 101) 0. 554 g of KOH reacts to give 1 g of KNO 3 (× 25. 0) 13. 9 g of KOH reacts to give 25. 0 g of KNO 3

Answers – Q 5 Mr of C 2 H 4 = (2 × 12) + (4 × 1) = 28 Mr of C 2 H 6 O = (2 × 12) + (6 × 1) + 16 = 46 28 g of C 2 H 4 reacts to give 46 g of C 2 H 6 O (÷ 46) 0. 609 g of C 2 H 4 reacts to give 1 g of C 2 H 6 O (× 60) 36. 5 g of C 2 H 4 reacts to give 60. 0 g of C 2 H 6 O

Answers – Q 6 Mr of 2 Fe 2 O 3 = 2 × ((2 × 56) + (3 × 16)) = 320 Mr of 4 Fe = 4 × 56 = 224 320 g of Fe 2 O 3 reacts to give 224 g of Fe (÷ 224) 1. 429 g of Fe 2 O 3 reacts to give 1 g of Fe (× 32. 0) 45. 7 g of Fe 2 O 3 reacts to give 32. 0 g of Fe