有关MOL的化学题

kbubble

有谁可帮我解这题?想了好久都想不出

2.10g of a mixture of potassium chlorate and potassium chloride were heated until the evolution of oxygen were complete.this residue weighed 1.62g.What is the percentage by mass of potassim chloride in the mixtuere?

答案是:41.7%

katami

The stoichiometry for the reaction ([] : subscript)

2KCL0[3](s) --> 2KCL (s) + 30[2] (g)

As oxygen is the only gaseous product, the 'weight' produced = 2.10 - 1.62 = 0.48g. The purpose is to find the equivalent mol of the oxygen produced and the equivalent mol/mass of KCL0[3] that produces it.

Equivalent mole of 0[2] = 0.48 / 32 = 0.015 mol
Equivalent mole of KCLO[3] = (0.015 x 2) / 3 = 0.01 mol
Equivalent mass of KCLO[3] = 0.01 mol  x (39.10 + 35.45 + (3x10)) = 1.2255g

KCL weight in mixture = 2.1 - 1.2255 = 0.8745g
Percentage = (0.875 / 2.1) x 100 = 41.66 %