Carbon monoxide is passed over heated iron(III) oxide in an open reaction tube. Carbon monoxide is in excess and all the iron(III) oxide reacts.
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) \mathrm{Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)} Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)The carbon monoxide and carbon dioxide leave the tube, so iron is the only solid remaining after the apparatus has cooled. The mass of this solid is 4.80 g less than the mass of iron(III) oxide before heating.
Relative atomic masses (ArA_rAr): O = 16.0 Fe = 55.8
What mass of iron(III) oxide was used?
4.80 g
11.2 g
16.0 g
6.86 g
134 exam-style questions on OCR GCSE Chemistry 3.1 Introducing chemical reactions, covering 3.1.1 Writing formulae and balanced chemical equations, 3.1.2 Formulae of compounds from ions, and state symbols, 3.1.3 Conservation of mass and observed changes in mass, 3.1.4 Half equations and balanced ionic equations, 3.1.5 The mole and the Avogadro constant, and 3.1.6 Stoichiometry, limiting reactants and reacting masses. Each one has a worked solution and a mark scheme showing where the marks go.