If my original sample contained 7.1 grams of potassium chlorate. How many moles of oxygen gas would I expect to have evolved? Hint: Convert grams to moles of potassium chlorate, then use the mole ratio in the balanced equation. Balance this equation KClO3 → KCl + O2(g)
I get .0869033048 when i do the calculations, want to make sure I did it right. 2KClO3 → 2KCl + 3O2(g)
7.1 g * (1 mol/122.55 g/mol) = .0579355465 mol * (3/2) = .0869033048
chemistry - DrBob222, Tuesday, April 1, 2014 at 10:34pm
Yes, your reasoning is flawless and your calculations are superb EXCEPT that you have only two significant figures in 7.1; therefore, you may not have more than 2 s.f. in the answer. So 0.086903 would become 0.087 rounded to 2 s.f. If you omitted a zero on 7.1 than you would be allowed 3 s.f. and the answer would be rounded to 0.0869 grams. Some profs are very picky about s.f.; some are not.