If 80.0 grams of methane were reacted with 368 grams of oxygen, how many grams of water would be formed (find limiter first).
As the problem notes this is a limiting reagent problem. I work these by solving two simple stoichiometry problems (simple meaning the two problems are not limiting.) Solve the first problem using one of the reactants, the second problem using the other reactant and you end up with two answers for the moles of H2O produced. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Here is an example of how to solve simple stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the limiting reactant in a chemical reaction, you need to compare the stoichiometric coefficients and the molar masses of the reactants.
Given:
Methane (CH4): 80.0 grams
Oxygen (O2): 368 grams
First, calculate the number of moles of each reactant:
Moles of Methane (CH4):
Molar mass of CH4 = molar mass of Carbon (12.01 g/mol) + 4 * molar mass of Hydrogen (1.01 g/mol)
Molar mass of CH4 = 12.01 g/mol + 4 * 1.01 g/mol = 16.05 g/mol
Number of moles of CH4 = Mass of CH4 / Molar mass of CH4
Number of moles of CH4 = 80.0 g / 16.05 g/mol ≈ 4.98 mol
Moles of Oxygen (O2):
Molar mass of O2 = 2 * molar mass of Oxygen (16.00 g/mol)
Molar mass of O2 = 2 * 16.00 g/mol = 32.00 g/mol
Number of moles of O2 = Mass of O2 / Molar mass of O2
Number of moles of O2 = 368 g / 32.00 g/mol ≈ 11.50 mol
Now, compare the moles of methane and oxygen:
The balanced chemical equation for the reaction of methane (CH4) with oxygen (O2) to form water (H2O) is:
CH4 + 2O2 -> CO2 + 2H2O
From the equation, we can see that the stoichiometric coefficient of methane (CH4) is 1, and the stoichiometric coefficient of oxygen (O2) is 2.
To determine which reactant is the limiting reactant, we compare the ratios of the moles of each reactant to their stoichiometric coefficients:
For methane (CH4):
Moles of CH4 / Stoichiometric coefficient of CH4 = 4.98 mol / 1 = 4.98 mol
For oxygen (O2):
Moles of O2 / Stoichiometric coefficient of O2 = 11.50 mol / 2 = 5.75 mol
The ratio of moles of CH4 to the stoichiometric coefficient is 4.98, while the ratio of moles of O2 to the stoichiometric coefficient is 5.75. Since the ratio for O2 is larger, it means that there is an excess of oxygen, and methane is the limiting reactant.
To calculate the amount of water formed, we need to use the stoichiometry of the balanced chemical equation.
From the balanced chemical equation:
1 mole of CH4 produces 2 moles of H2O
Since 4.98 moles of CH4 are present and methane is the limiting reactant, we can calculate the moles of water formed:
Moles of H2O = 2 * Moles of CH4
Moles of H2O = 2 * 4.98 mol = 9.96 mol
Finally, calculate the mass of water formed:
Mass of H2O = Moles of H2O * Molar mass of H2O
Molar mass of H2O = molar mass of Hydrogen (1.01 g/mol) * 2 + molar mass of Oxygen (16.00 g/mol) = 18.02 g/mol
Mass of H2O = 9.96 mol * 18.02 g/mol ≈ 179.61 grams
Therefore, approximately 179.61 grams of water would be formed.