How many grams of oxygen are needed to react completely with 10 moles of methane(ch4)?

Question

using this reaction.. ch4+2o2 --> co2+2h2o

Answer 1

since u have a balanced equation, and u want 10 moles of (CH4), multiply all reactants and products by 10, so u get:
10(CH4) + 20 (O2) --> 10(CO2) + 20(H2O)
this means u need 20 moles of (O2), since the atomic weight of oxygen is 16, and u have 2 atoms of oxygen in each (O2) molecule, u need 20*16*2=640 grams of oxygen...

Answer 2

First you stability the equation 2 H2 + O2 ------ 2 H2O so 2 moles of hydrogen react with ONE mole of oxygen to offer 2 moles of water. on condition that H2 and O2 react in a 2:a million ratio then 10 moles of H2 require 5 moles of O2 to thoroughly react. From 10 moles of H2 and 5 moles of oxygen one gets 10 moles of water. on condition that each and every mole of water has a mass of 18 grams in line with mole 10 moles would have a mass of 10 X 18 = a hundred and eighty grams for Hydrogen 10 moles = 10 X 2grams/mole = 20 grams

Answer 3

You will need to use the molar ratio from the balanced equation:

10 mol CH4 X [(2 mol O2)/(1 mol CH4)] X [(32 grams O2)/(1 mol O2)]

Notice all the units cancel out except grams of O2. Multiply all numbers on top and then divide by the sum of the numbers on the bottom.

640 grams of O2

Hope this helped! Good luck!!

Answer 4

From the balanced equation, two moles of O2 are required per mole of methane, at 32 grams each. Hence, 640 grams.