1. The problem statement, all variables and given/known data A compound containing element A and oxygen has a mole ratio of A:O = 2:3. If 8.0 grams of the oxide contains 2.4 grams of oxygen... a) what is the atomic weight of A? b) what is the weight of one mole of the oxide? c) what is the theoretical weight of the oxide formed when 28 g of A is heated in excess oxygen? d) what is the % yield if 38 grams was produced from 28 grams of A? 2. Relevant equations .... 3. The attempt at a solution I didn't have any because I barely understood what to do :( can someone please guide me on how to solve this? It'll be of great help ^^
Start with the oxygen. How many moles of oxygen is equivalent to that 2.4 grams of oxygen from the compound?
8.0 - 5.6 = 2.4 gO 2.4 gO x 1 mole O / 16 gO = 0.15 moles O A:O = 2:3 A:0.15 = 2:3 [tex]\frac{A}{0.15}[/tex] x [tex]\frac{2}{3}[/tex] 3A = (0.15)(2) 3A = 0.30 A = 0.1 moles Oxygen a) AW = g/mole AW = 5.6 gO / 0.1 moles O AW = 56 g/mole b) 2:3 A_{2}O_{3} = (56)(2) + (16)(3) = 160 g Is this correct? I'm sure with my a & b but with c and d, I don't really get how that happened :( I asked for help in solving this, and this was the solution he gave.
oops! I'm very sorry for that the one in my paper was 0.1 moles of A :D haha xD thank you! um, but the part that I'm really confused about is this one ... 4A + 3O_{2} -> 2 A_{2}O_{3} if we take the balancing off.. A + O_{2} -> A_{2}O_{3} Where did we get A + O_{2} ? Is this common knowledge? because all I understood was A_{2}O_{3} since we had a mole ratio of 2:3 right? :D
Not sure what your problem is. There are many ways of producing oxides, but direct reaction between the element and oxygen is the simplest one (even if technically not always possible). Even if it is not possible, you were told in the question that 38 grams of A_{2}O_{3} were produced from 28 grams of A - no matter what the real reaction was, molar ratio of the A and A_{2}O_{3} will be always the same. Then A_{2}O_{3} was given, O[ub]2[/sub] is a common knowledge, A is just the simplest way of approaching the problem.