The combustion of a sample of butane, c4h10 (lighter fluid), produced 2.46 grams of water.

2 c4h10 + 13 o2 > 8 co2 + 10 h2o

(a) how many moles of water formed?

(b) how many moles of butane burned?

(c) how many grams of butane burned?

(d) how much oxygen was used up in moles? in grams?
chem

A. 2.46 g H2O × 1 mol H2O/18.02 g H2O = 0.137 mol H2O

b. 2.46 g H2O × 1 mol H2O/18.02 g H2O × 2 mol C4H10/10 mol H2O = 0.0273 mol C4H10

c. 2.46 g H2O × 1 mol H2O/18.02 g H2O × 2 mol C4H10/10 mol H2O × 58.12 g C4H10/1 mol C4H10 = 1.59 g C4H10

d. 
moles of O2
2.46 g H2O × 1 mol H2O/18.02 g H2O × 13 mol O2/10 mol H2O = 0.177 mol O2

Grams of O2
2.46 g H2O × 1 mol H2O/18.02 g H2O × 13 mol O2/10 mol H2O × 32.00 g O2/1 mol O2 = 5.68 g O2

This would be canberra.

Yes she is correct. the answer is completely correct