The reaction between magnesium and oxygen is: 2 Mg ( s ) + O 2 ( g ) → 2 MgO ( s ) Write a word equation for the reaction.

The reaction between magnesium and oxygen is: 2 Mg ( s ) + O 2 ( g ) → 2 MgO ( s ) Two moles of magnesium react with one mole of oxygen to produce two moles of magnesium oxide. The word equation for the reaction is as follows: Magnesium + Oxygen → Magnesium oxide

The reaction between magnesium and oxygen is: 2 Mg ( s ) + O 2 ( g ) → 2 MgO ( s ) How many moles of magnesium atoms react with 1 mole of oxygen molecules?

The reaction between magnesium and oxygen is: 2 Mg ( s ) + O 2 ( g ) → 2 MgO ( s ) From the reaction, it is clear that two moles of magnesium react with one mole of oxygen to produce two moles of magnesium oxide.

The reaction between magnesium and oxygen is: 2 Mg ( s ) + O 2 ( g ) → 2 MgO ( s ) The A r values are: Mg = 24 , O = 16 . How many grams of oxygen react with: 48 g of magnesium?

The reaction between magnesium and oxygen is: 2 Mg ( s ) + O 2 ( g ) → 2 MgO ( s ) From the reaction, it is clear that two moles of magnesium react with one mole of oxygen to produce two moles of magnesium oxide. Number of moles of magnesium in its 48 g = 48 g 24 g = 2 mol Therefore, one mole of oxygen will react with 48 g of magnesium. Since A r of oxygen is 16 u . Hence, its M r is 32 u . The mass of one mole of oxygen is 32 g .

The reaction between magnesium and oxygen is: 2 Mg ( s ) + O 2 ( g ) → 2 MgO ( s ) The A r values are: Mg = 24 , O = 16 . How many grams of oxygen react with: 12 g of magnesium?

Number of moles of magnesium in its 12 g = 12 g 24 g = 0 . 5 mol The reaction between magnesium and oxygen is: 2 Mg ( s ) + O 2 ( g ) → 2 MgO ( s ) From the reaction, it is clear that two moles of magnesium react with one mole of oxygen to produce two moles of magnesium oxide. Therefore, 0 . 5 mol or 12 g of magnesium will react with 0 . 25 mol of oxygen. Gram M r of oxygen is 32 g . Mass of 0 . 25 mol of oxygen is 0 . 25 mol × 32 g = 8 g

Copper carbonate breaks down on heating, like this: CuCO 3 ( s ) → CuO ( s ) + CO 2 ( g ) Write a word equation for the reaction.

Copper carbonate breaks down on heating, like this: CuCO 3 ( s ) → CuO ( s ) + CO 2 ( g ) In the reaction, one mole of reactant decomposes into one mole of each of the products. The word equation for this reaction is as follows: copper carbonate → copper oxide + carbon dioxide

Copper carbonate breaks down on heating, like this: CuCO 3 ( s ) → CuO ( s ) + CO 2 ( g ) Find the mass of 1 mole of each substance taking part in the reaction. A r : Cu = 64 , C = 12 , O = 16

Mass of one mole of any substance is the molar mass of the substance which is equal to its molecular mass represented in grams. Since, only copper carbonate is taking part in this reaction, it is the only reactant. Mass of one mole of copper carbonate = 64 u + 12 u + 3 × 16 u = 124 u

Copper carbonate breaks down on heating, like this: CuCO 3 ( s ) → CuO ( s ) + CO 2 ( g ) When 31 g of copper (ll) carbonate is used how many grams of carbon dioxide is formed? A r : Cu = 64 , C = 12 , O = 16

Mass of one mole of copper carbonate = 64 g + 12 g + 3 × 16 g = 124 g Number of moles of copper carbonate in its 31 g = 31 g 124 g = 0 . 25 mol CuCO 3 ( s ) → CuO ( s ) + CO 2 ( g ) From the reaction it is clear that 1 mol of copper carbonate produces 1 mol of carbon dioxide. Therefore, 0 . 25 mol of copper carbonate produces 0 . 25 mol of carbon dioxide. Mass of 0 . 25 mol of carbon dioxide 0 . 25 mol × 44 g = 11 g

Copper carbonate breaks down on heating, like this: CuCO 3 ( s ) → CuO ( s ) + CO 2 ( g ) When 31 g of copper (ll) carbonate is used, what mass of solid remains after heating? A r : Cu = 64 , C = 12 , O = 16

Mass of one mole of copper carbonate = 64 g + 12 g + 3 × 16 g = 124 g Number of moles of copper carbonate in its 31 g = 31 g 124 g = 0 . 25 mol CuCO 3 ( s ) → CuO ( s ) + CO 2 ( g ) From the reaction it is clear that 1 mol of copper carbonate produces 1 mol of carbon dioxide. Therefore, 0 . 25 mol of copper carbonate produces 0 . 25 mol of carbon dioxide and 0 . 25 mol of solid remains after heating. Mass of one mole of copper oxide = 64 g + 16 g = 80 g Mass of 0 . 25 mol of copper oxide 0 . 25 mol × 80 g = 20 g

E M B I B E

Complete Chemistry for Cambridge IGCSE® Second Edition>Chapter 6 - Using Moles>Q

RoseMarie Gallagher and Paul Ingram Solutions for Chapter: Using Moles, Exercise 2: Q

Author:RoseMarie Gallagher & Paul Ingram

RoseMarie Gallagher Chemistry Solutions for Exercise - RoseMarie Gallagher and Paul Ingram Solutions for Chapter: Using Moles, Exercise 2: Q

Attempt the free practice questions on Chapter 6: Using Moles, Exercise 2: Q with hints and solutions to strengthen your understanding. Complete Chemistry for Cambridge IGCSE® Second Edition solutions are prepared by Experienced Embibe Experts.