'O' & 'A' Level Qns.
Qn 1.
When 1.01g sample of potassium nitrate(V) is heated above its melting point, oxygen gas is evolved. The mass of the sample decreases by 0.16g. Construct a balanced equation for this decomposition reaction, name the residue and (for 'A' level students) draw its structural formula.
Q2.
Sulfur and chlorine can react together to form disulfur dichloride. When 1.00g of disulfur dichloride was reacted with water, 0.36g of a yellow precipitate of sulfur was formed together with a solution containing a mixture of sulfuric(IV) acid (a.k.a. sulfurous acid), and hydrochloric acid. Given that the most common isotope of sulfur is S8, use the above data to deduce the balanced equation for the reaction between disulfur dichloride and water. (For 'A' level students) draw the structural formulae of disulfur dichloride and sulfuric(IV) acid.
Solutions (partial).
Qn1. Since 1 mole of potassium nitrate(V) decomposes to give 0.5 mole of oxygen gas, the residue is potassium nitrate(III) {stock name} or potassium nitrite {latin name}.
For the nitrate(III) ion (aka nitrite ion), uninegative formal charge on a singly bonded O atom; central N atom has 1 lone pair, 3 bond pairs; electron geometry tetrahedral, ionic geometry trigonal pyramidal.
Qn2. Since 1 mole of disulfur dichloride undergoes hydrolysis to give 1.5 moles of S(s), or 1.5/8 moles of S8(s), the coefficients of the balanced equation are : 2, 3, 1, 4, 3/8
For disulfur dichloride, the 2 central S atoms have 2 lone pairs and 2 bond pairs; electron geometry tetrahedral, molecular geometry v-shape.
For sulfuric(IV) acid {stock name} or sulfurous acid {latin name}, the central S atom has 1 lone pair, 4 bond pairs (S can expand its octet using its vacant and energetically accessible 3d orbitals); electron geometry tetrahedral, molecular geometry trigonal pyramidal.