Hydrogen reacts with chlorine to form hydrogen chloride. All three substances are at −100 ∘C-100\,^\circ\mathrm{C}−100∘C.
H2(…)+Cl2(…)→2HCl(…) \mathrm{H_{2}(\ldots) + Cl_{2}(\ldots) \rightarrow 2HCl(\ldots)} H2(…)+Cl2(…)→2HCl(…)| Substance | Melting point (∘C^\circ\mathrm{C}∘C) | Boiling point (∘C^\circ\mathrm{C}∘C) |
|---|---|---|
| Hydrogen | -259 | -253 |
| Chlorine | -101 | -34 |
| Hydrogen chloride | -114 | -85 |
What is the equation with the correct state symbol for every substance at −100 ∘C-100\,^\circ\mathrm{C}−100∘C?
H2(g)+Cl2(l)→2HCl(l)\mathrm{H_{2}(g) + Cl_{2}(l) \rightarrow 2HCl(l)}H2(g)+Cl2(l)→2HCl(l)
H2(g)+Cl2(s)→2HCl(s)\mathrm{H_{2}(g) + Cl_{2}(s) \rightarrow 2HCl(s)}H2(g)+Cl2(s)→2HCl(s)
H2(l)+Cl2(l)→2HCl(l)\mathrm{H_{2}(l) + Cl_{2}(l) \rightarrow 2HCl(l)}H2(l)+Cl2(l)→2HCl(l)
H2(g)+Cl2(g)→2HCl(g)\mathrm{H_{2}(g) + Cl_{2}(g) \rightarrow 2HCl(g)}H2(g)+Cl2(g)→2HCl(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.