If 8.6 l of o2 reacted with excess h2 at stp, what is volume of the gaseous water collected? @h2 (g) + o2 (g) = 2h2 o (g)

Note that this is occurring at STP, where 22.4L of any gas is equal to 1mol of that gas.

First, convert the liters of O₂ to moles of O₂ using the conversion factor 22.4LO₂ = 1molO₂.
8.6LO₂ × 1molO₂/22.4LO₂
= 8.6/22.4
≈ 0.3839molO₂

Next, convert moles of O₂ to moles of H₂O. In the balanced equation, the coefficients show that there are 2 moles of H₂O for every mole of O₂. So, use the conversion factor 1molO₂ = 2molH₂O.
0.3839molO₂ × 2molH₂O/1molO₂
= 0.3839 × 2
= 0.7678molH₂O

Finally, convert the moles of H₂O to liters of H₂O using the same conversion factor from before, 22.4LH₂O = 1molH₂O.
0.7678molH₂O × 22.4LH₂O/1molH₂O
= 0.7678 × 22.4
≈ 17LH₂O

So, the answer is 17 liters of gaseous water is collected! Note that its rounded to 17 because the measurement given in the problem has 2 sig figs. Hope that helps! :)