how many grams of water can be formed when 28.7 grams of CH4 reacts with 45.8 grams of O2
CH4 + 2O2 ==> CO2 + 2H2O
Here is a worked example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the grams of water formed when CH4 reacts with O2, we need to first determine the balanced chemical equation for the reaction.
The balanced equation for the reaction between CH4 (methane) and O2 (oxygen) is:
CH4 + 2O2 -> CO2 + 2H2O
This equation tells us that for every 1 mole of CH4, we require 2 moles of O2 to form 2 moles of water (H2O).
To find out how many moles of CH4 and O2 we have, we can use the molar masses of the elements involved.
The molar mass of CH4 (carbon + 4 hydrogens) = 12.01 g/mol + (4 * 1.01 g/mol) = 16.05 g/mol
The molar mass of O2 (2 oxygens) = 2 * 16.00 g/mol = 32.00 g/mol
Next, we determine the number of moles of CH4 and O2:
moles of CH4 = mass of CH4 / molar mass of CH4
= 28.7 g / 16.05 g/mol
≈ 1.79 mol
moles of O2 = mass of O2 / molar mass of O2
= 45.8 g / 32.00 g/mol
≈ 1.43 mol
From the balanced equation, we see that 1 mole of CH4 reacts with 2 moles of O2 to form 2 moles of H2O.
Since we have 1.79 moles of CH4, we can expect the amount of H2O formed to be twice that amount:
moles of H2O formed = (moles of CH4) * 2
= 1.79 mol * 2
= 3.58 mol
Finally, we can calculate the mass of water formed:
mass of H2O = moles of H2O * molar mass of H2O
= 3.58 mol * (2 * 1.01 g/mol)
≈ 7.24 g
Therefore, approximately 7.24 grams of water can be formed when 28.7 grams of CH4 reacts with 45.8 grams of O2.