What mass of water would result given complete combustion of a #44.0*g# mass of propane? Chemistry Stoichiometry Stoichiometry 1 Answer anor277 Jun 5, 2016 Approx. #72*g#. Explanation: #C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O# #"Moles of propane"# #=# #(44.0*g)/(44.10*g*mol^-1)# #~=# #1*mol#. Given the balanced equation, how many moles of water will result, and how much mass does this constitute? Answer link Related questions How do you solve a stoichiometry problem? What is stoichiometry? Question #93ac6 Why do we study stoichiometry? How many grams of NaOH is produced from #1.20 x 10^2# grams of #Na_2O#? #Na_2O + H_2O -> 2NaOH# How many grams of Na2O are required to produce 1.60 x 102 grams of NaOH? Na2O + H2O ---> 2 NaOH What mass of iron is needed to react with 16.0 grams of sulfur? 8 Fe + S8 ---> 8 FeS According to 8 Fe + S8 ---> 8 FeS How many grams of FeS are produced? 12.00 moles of NaClO3 will produce how many grams of O2? 2 NaClO3 ---> 2 NaCl + 3 O2 How many grams of NaCl are produced when 80.0 grams of O2 are produced? 2 NaClO3 ---> 2 NaCl + 3 O2 See all questions in Stoichiometry Impact of this question 1483 views around the world You can reuse this answer Creative Commons License