Читать книгу Foundations of Chemistry - Philippa B. Cranwell - Страница 44

1 Mg + O2 → MgOThis requires 2 atoms of O on the right‐hand side: Mg + O2 → 2MgO.And the left‐hand side now requires 2 atoms of Mg: 2Mg + O2 → 2MgO.

2 NaOH + H2SO4 → Na2SO4 + H2OThe left‐hand side requires 2 atoms of Na: 2NaOH + H2SO4 → Na2SO4 + H2O.Now the right‐hand side needs 2 moles of H2O: 2NaOH + H2SO4 → Na2SO4 + 2H2O.

3 CO + NO → CO2 + N2The left‐hand side needs 2 atoms of N: CO + 2NO → CO2 + N2.The left‐hand side now has 3 atoms of O and the right‐hand side has 2, so increase the number of moles of CO2 on the right: CO + 2NO → 2CO2 + N2.Now the left‐hand side needs one more atom of C and one more atom of O: 2CO + 2NO → 2CO2 + N2.

4  C2H5OH + O2 → CO2 + H2OWhen a hydrocarbon is burnt in oxygen, the products are carbon dioxide and water. First balance the number of atoms of carbon on each side of the equation: C2H5OH + O2 → 2CO2 + H2O.Now balance the number of atoms of hydrogen by increasing the number of moles of water to 3: C2H5OH + O2 → 2CO2 + 3H2O.Next balance the number of atoms of oxygen by adding 3 moles of O2 on the left‐hand side: C2H5OH + 3O2 → 2CO2 + 3H2O.