O-Level Chemical Calculations: worked solution

Question

Balance each equation below. (i) __Zn + __HCl -> __ZnCl2 + __H2 (ii) __CuCO3 + __HCl -> __CuCl2 + __CO2 + __H2O (iii) __NaOH + __H2SO4 -> __Na2SO4 + __H2O

Rule: you may only change the big numbers written in front of the formulae. Never change a small subscript inside a formula, because that would change the substance itself. Method: count each type of atom on both sides, then even them up and check again. (i) __Zn + __HCl -> __ZnCl2 + __H2 The right-hand side has 2 Cl (in ZnCl2) and 2 H (in H2), but the left has only 1 of each. Put a 2 in front of HCl to supply both. Zn + 2HCl -> ZnCl2 + H2 Check: Zn 1 = 1, H 2 = 2, Cl 2 = 2. (ii) __CuCO3 + __HCl -> __CuCl2 + __CO2 + __H2O CuCl2 needs 2 Cl, and each HCl supplies only 1, so put a 2 in front of HCl. CuCO3 + 2HCl -> CuCl2 + CO2 + H2O Check: Cu 1 = 1, C 1 = 1, Cl 2 = 2, H 2 = 2. Oxygen: left has 3 (in CuCO3); right has 2 (in CO2) + 1 (in H2O) = 3. (iii) __NaOH + __H2SO4 -> __Na2SO4 + __H2O Na2SO4 needs 2 Na, so put a 2 in front of NaOH. Now the left has H = 2 (from 2NaOH) + 2 (from H2SO4) = 4, so put a 2 in front of H2O. 2NaOH + H2SO4 -> Na2SO4 + 2H2O Check: Na 2 = 2, S 1 = 1, H 4 = 4. Oxygen: left has 2 (in 2NaOH) + 4 (in H2SO4) = 6; right has 4 (in Na2SO4) + 2 (in 2H2O) = 6.