if 8.50grams CH4 reacts with 15.9grams of oxygen gass which of the prdocut is the excess and which is the limiting.
One of the REACTANTS is the limiting reagent. A product can't be a limiting reagent.
sorry i mean which reactant
See your other post below. I detailed how to do it. Same kind of problem.
To determine which substance is in excess and which is the limiting reactant, we need to compare the amounts of CH4 and O2 in the reaction.
Step 1: Write and balance the chemical equation:
CH4 + 2O2 → CO2 + 2H2O
Step 2: Calculate the molar masses:
CH4 = 12.01 + (4x1.01) = 16.05 g/mol
O2 = 2x16.00 = 32.00 g/mol
Step 3: Convert the given masses of CH4 and O2 to moles using their respective molar masses:
moles of CH4 = 8.50 g CH4 / 16.05 g/mol = 0.529 mol CH4
moles of O2 = 15.9 g O2 / 32.00 g/mol = 0.497 mol O2
Step 4: Determine the stoichiometric ratio between CH4 and O2. From the balanced equation, we can see that the ratio is 1:2. This means that 1 mole of CH4 reacts with 2 moles of O2.
Step 5: Calculate the moles of O2 needed to react with all the CH4:
moles of O2 needed = 0.529 mol CH4 x (2 mol O2 / 1 mol CH4) = 1.058 mol O2
Step 6: Compare the moles of O2 needed (1.058 mol) with the moles of O2 available (0.497 mol). Since we have fewer moles of O2 available than needed, O2 is the limiting reactant.
Step 7: Calculate the moles of the excess reactant (CH4):
moles of CH4 left = moles of CH4 initially - moles of CH4 reacted
moles of CH4 left = 0.529 mol CH4 - 0.497 mol CH4
moles of CH4 left = 0.032 mol CH4
Based on the calculations, O2 is the limiting reactant because there is less O2 available than needed. CH4 is the excess reactant because there is some remaining after the reaction is complete.