text: how to find the volume of h_2o (g)? considering.... that h_2o (g) measured at stp is produced by the combustion of 4.00 (g )of natural gas according to tge following equations. ch_4 (g) + 2o_2 (g) ---> co_2 (g) + 2h_2o (g) the answer is 11.2 l  first of all, i know that i have to do some grams to mole conversation in order to find n. then plug in stp p and t, r, and n into my equation... pv=nrt however, in no way i am getting 11.2 l as the final answer  i have tried various reasons and i am sure that i am making a  slight mistake with the moles. please show all work in order to see where i was wrong. thanks. 4 grams of ch_4 which is 4/16(molar mass)=.25 moles multiply.25 moles of ch_4 by two because of the two moles of water. .5moles of water and at stp volume of one mole of any gas is 22.4 l so multiply .5moles by 22.4 and get 11.2