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.
2.5 g of zinc are reacted with 9.8 g of hydrogen chloride. Which is the limiting reactant?
Zn + 2HCl → ZnCl₂ + H₂. Calculate the number of moles for each reactant. Zn: n = m/Mᵣ = 2.5/65 = 0.04. HCl: n = m/Mᵣ = 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.
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 → Na₂S. 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.
Calculate the limiting reactant in the reaction of 5.6 g of iron(II) oxide with 6.7 g of hydrogen.
FeO + H₂ → Fe + H₂O. Moles of FeO = 5.6/72 = 0.08. Moles of H₂ = 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.
14.5 g of ammonia are reacted with 7.2 g of oxygen to produce nitric oxide and water. Which reactant is in excess?
4NH₃ + 5O₂ → 4NO + 6H₂O. Moles of NH₃ = 14.5/17 = 0.85. Moles of O₂ = 7.2/32 = 0.225. Molar ratio = 4 : 5 = 0.85 : 1.06. There are only 0.225 moles of O₂, so NH₃ 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?
2Mg + O₂ → 2MgO. Moles of Mg = 12/24 = 0.5. Moles of O₂ = 3.2/32 = 0.1. Molar ratio = 2 : 1 = 0.5 : 0.25. There are only 0.1 moles of O₂, so O₂ is the limiting reactant and Mg is in excess.
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