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

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.

To calculate the number of moles of oxygen gas evolved, you should follow the steps you outlined in your question.

1. Convert grams of potassium chlorate (KClO3) to moles:
The molar mass of potassium chlorate (KClO3) is 122.55 g/mol.
Given: 7.1 grams of KClO3
Calculation: 7.1 g * (1 mol/122.55 g/mol) = 0.0579355465 mol of KClO3

2. Use the balanced equation to determine the mole ratio between KClO3 and O2:
Balanced equation: 2KClO3 → 2KCl + 3O2(g)
According to the equation, for every 2 moles of KClO3, 3 moles of O2 are produced.

3. Calculate the number of moles of O2 evolved:
Calculation: 0.0579355465 mol of KClO3 * (3/2) = 0.0869033048 mol of O2

You have correctly calculated that you would expect to have approximately 0.0869033048 moles of oxygen gas evolved.

Therefore, your answer of 0.0869033048 is correct. Well done!