OpenStudy (anonymous):

How many L of oxygen, at STP, are required to form 10.5g of water, 2H2 + O2 --> 2H20

1 year ago
OpenStudy (jebonna):

The equation will help you alot in this question. Firstly what we need to do is find the moles of water. We can do this by using this equation: mass/molar mass (Mass = 10.5g, molar mass = 18g mol-1) Lets plug in our values: 10.5g/18g mol-1 = 0.583mol (3.sig.fig) Now we have the moles of water, we can deduce the moles to find the number of moles for oxygen. The moles of water is equal to TWO moles of water (as you can see on the equation, there is a 2 in front of H2O), and we need the answer for the moles equal to ONE mole of O2 (as you can see in front of the O2 there is no number, meaning there is just 1) So to make our answer for one mole of a substance instead of two, we just have to divide our answer by 2 to get the amount of moles for 1. 0.583mol/2 = 0.292mol of ONE mole of O2 (3.sig.fig) Because 1 mol = 22.4 L, now to find the volume of the gas in litres, we need to do this equation: moles x 22.4L Lets plug in the values: 0.292mol x 22.4L = 6.54L of O2 (3.sig.fig) So your answer is: 6.54L of O2 is needed to make 10.5g of H2O I hope this helps :)

1 year ago