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Formulae and equations for SQA Higher Chemistry

Formulae and equations

This page covers the following topics:

1. Limiting reactants

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

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.

4

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?

5

Which reactant is in excess when 12 g of magnesium are reacted with 3.2 g of oxygen to produce magnesium oxide?

2Mg + Oโ‚‚ โ†’ 2MgO. Moles of Mg = 12/24 = 0.5. Moles of Oโ‚‚ = 3.2/32 = 0.1. Molar ratio = 2 : 1 = 0.5 : 0.25. There are only 0.1 moles of Oโ‚‚, so Oโ‚‚ is the limiting reactant and Mg is in excess.

Which reactant is in excess when 12 g of magnesium are reacted with 3.2 g of oxygen to produce magnesium oxide?

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