Asked by bekah
                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
            
        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
Answers
                    Answered by
            DrBob222
            
    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. 
    
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