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Formulae and equations for AQA GCSE Chemistry

Formulae and equations

This page covers the following topics:

1. Balancing equations
2. Limiting reactants

A balanced equation is a model of a chemical reaction that shows the formulae of the reactants and the products. In order for an equation to be balanced, the number of atoms must be equal on both sides of the equation.

Balancing equations

The limiting reactant is the first reactant to be used up in a chemical reaction. When the limiting reactant is used up, the reaction stops and no more of the product can be formed. Therefore, the mass of the product is dependent on the mass of the limiting reactant. The left-over reactant is referred to as the reactant in excess.

Limiting reactants

1

2.5 g of zinc are reacted with 9.8 g of hydrogen chloride. Which is the limiting reactant?

Zn + 2HCl โ†’ ZnClโ‚‚ + Hโ‚‚. Calculate the number of moles for each reactant. Zn: n = m/Mแตฃ = 2.5/65 = 0.04. HCl: n = m/Mแตฃ = 9.8/36 = 0.3. Consider the ratio of the reactants: 1 mol Zn : 2 mol HCl. 0.15 mol Zn : 0.3 mol HCl. As there are only 0.04 mol of Zn, Zn is the limiting reactant.

2.5 g of zinc are reacted with 9.8 g of hydrogen chloride. Which is the limiting reactant?

2

5.7 g of sodium are reacted with 6.1 g of sulfur. Which is the limiting reactant and which reactant is in excess?

Na is the limiting reactant and S is in excess

2Na + S โ†’ Naโ‚‚S. Moles of Na = 5.7/23 = 0.25. Moles of S = 6.1/32 = 0.19. Molar ratio = 2 : 1 = 0.38 : 0.19. There are only 0.25 moles of Na, so Na is the limiting reactant and S is in excess.

5.7 g of sodium are reacted with 6.1 g of sulfur. Which is the limiting reactant and which reactant is in excess?

3

Ethane reacts with oxygen to produce carbon dioxide and water. Write the balanced equation for this reaction.

2Cโ‚‚Hโ‚† + 7Oโ‚‚ โ†’ 4COโ‚‚ + 6Hโ‚‚O

Ethane reacts with oxygen to produce carbon dioxide and water. Write the balanced equation for this reaction.

4

Calculate the limiting reactant in the reaction of 5.6 g of iron(II) oxide with 6.7 g of hydrogen.

FeO + Hโ‚‚ โ†’ Fe + Hโ‚‚O. Moles of FeO = 5.6/72 = 0.08. Moles of Hโ‚‚ = 6.7/2 = 3.35. Molar ratio = 1 : 1 = 3.35 : 3.35. There are only 0.08 moles of FeO, so FeO is the limiting reactant.

Calculate the limiting reactant in the reaction of 5.6 g of iron(II) oxide with 6.7 g of hydrogen.

5

14.5 g of ammonia are reacted with 7.2 g of oxygen to produce nitric oxide and water. Which reactant is in excess?

4NHโ‚ƒ + 5Oโ‚‚ โ†’ 4NO + 6Hโ‚‚O. Moles of NHโ‚ƒ = 14.5/17 = 0.85. Moles of Oโ‚‚ = 7.2/32 = 0.225. Molar ratio = 4 : 5 = 0.85 : 1.06. There are only 0.225 moles of Oโ‚‚, so NHโ‚ƒ is in excess.

14.5 g of ammonia are reacted with 7.2 g of oxygen to produce nitric oxide and water. Which reactant is in excess?

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