Question

Part A Write a balanced equation to show the reaction of sulfurous acid with rubidium hydroxide to form water and rubidium sulfite. OHSO3(aq) + RbOH(aq) - H200) + RbSO3(aq) H2S(aq) + 2 RbOH(aq) + 2 H200) + Rb 2S(aq) H2SO4(aq) + RbOH(aq) + H200) + Rb 2S04(aq) HSO4(aq) + RbOH(aq) - H2O(0) + RbSO4(aq) H2SO3(aq) + 2 RbOH(aq) -- 2 H200 + Rb 2503(aq) Submit Request Answer

Answer 1

A) Balanced equation is: Hz 503 109) + 2 R504 (aq) Left hand side (LAS) H shows = 1x2+2x1 = 4 saham = 1 0 atoms = 1x3 + 2x1 = 5 → 2420 (4) + R by S03 (aq) Right hand side (RMS) H atoms = 2x2 = 4 salom = 1 O atoms = 2x1 + 1X3:5 Rb atoms = 1x2= 2 Rb atoms : 281=2 Since; LHS = RHS (equation is balanced) * BU3 (9) + 3H2011) + 3Hellaq) + BloMlz Cuq) 1 mol of Bla produces 3 moles of Mel Modes of HU - 10.og 70.2743 mol 36.49gtmold moles of Bly needed = 1/2 x moles of Hel = x 0.2743 med = 0.0914 mol * 3Nog (9) + H2O(l) → 2HNO3(aq) + NO (9) 3 moles of Nog produces 2 mol of ANO3 1 mol of Noz produces & 2 mul of HNO3 3.75 mol of Noy will produce = ? nad of HNO3X3.75 = 2.50 moles HNO3