In one experiment 10.0g NH3 and 10.0g CO2 were combined and allowed to react. Determine how many grams of CH4N2O could be produced and how much of the reactant in excess will remain at the end of the reaction.
Chemical equation: 2NH3 + CO2 = CH4N2O + H2O
2NH3 + CO2 = CH4N2O + H2O
mols NH3 = grams/molar mass = 10/17 = 0.589
mols CO2 = 10/44 = 0.227
mols CH4N2O produced if NH3 is the limiting reagent (LR).
0.589 mols NH3 x (1 mol CH4N2O/2 mol NH3) = 0.294
mols CH4N2O produced if CO2 is the LR.
0.227 mols CO2 x (1 mol CH4N2O/1 mol CO2) = 0.227
The small number always wins in LR problems; therefore, CO2 is the LR and NH3 is the ER (excess reagent).
grams CH4N2O produced = mols CH4N2O produced x molar mass CH4N2O.
NH3 is the ER. How much is used? That's 0.227 mols CO2 x (2 mol NH3/1 mol CO2) = 0.227 x 2 = 0.454. So how much NH3 remains? That's 0.589 - 0.454 = ? mols NH3 remaining. Convert to grams if needed. Be sure and check my arithmetic.
To determine how many grams of CH4N2O can be produced and how much of the reactant will remain in excess, we need to follow these steps:
1. Calculate the number of moles of NH3 and CO2:
Moles of NH3 = mass of NH3 / molar mass of NH3
Moles of CO2 = mass of CO2 / molar mass of CO2
2. Determine the stoichiometric ratio between NH3 and CH4N2O:
From the balanced chemical equation, we can see that 2 moles of NH3 react to produce 1 mole of CH4N2O.
3. Determine the limiting reactant:
The limiting reactant is the one that is completely consumed, limiting the amount of product that can be formed. To identify the limiting reactant, compare the moles of NH3 and CO2 using the stoichiometric ratio. The reactant with fewer moles is the limiting reactant.
4. Calculate the theoretical yield of CH4N2O:
The theoretical yield is the maximum amount of product that can be formed from the limiting reactant. Multiply the moles of the limiting reactant by the stoichiometric ratio to find the moles of CH4N2O that can be produced.
Moles of CH4N2O = Moles of limiting reactant (NH3) / 2
5. Convert the moles of CH4N2O to grams:
Multiply the moles of CH4N2O by the molar mass of CH4N2O to obtain the mass in grams.
6. Calculate the excess reactant:
The excess reactant is the reactant that is not completely consumed. Subtract the moles of limiting reactant used in the reaction from the initial moles of the excess reactant.
Let's now perform the calculations:
1. Calculate the moles of NH3:
Molar mass of NH3 = (1x1.01) + (3x1.01) = 17.03 g/mol
Moles of NH3 = 10.0 g NH3 / 17.03 g/mol = 0.588 mol NH3
2. Calculate the moles of CO2:
Molar mass of CO2 = (1x12.01) + (2x16.00) = 44.01 g/mol
Moles of CO2 = 10.0 g CO2 / 44.01 g/mol = 0.227 mol CO2
3. Determine the limiting reactant:
Comparing the moles of NH3 and CO2, we see that NH3 is the limiting reactant since it has fewer moles.
4. Calculate the moles of CH4N2O:
Moles of CH4N2O = 0.588 mol NH3 / 2 = 0.294 mol CH4N2O
5. Calculate the mass of CH4N2O:
Molar mass of CH4N2O = (1x12.01) + (4x1.01) + (2x14.01) + (1x16.00) = 90.08 g/mol
Mass of CH4N2O = 0.294 mol CH4N2O * 90.08 g/mol = 26.48 g
6. Calculate the excess reactant:
Moles of excess CO2 = Initial moles of CO2 - moles of CO2 used
Moles of CO2 used = (0.588 mol NH3 / 2) * (1 mol CO2 / 2 mol NH3) = 0.147 mol CO2
Moles of excess CO2 = 0.227 mol CO2 - 0.147 mol CO2 = 0.080 mol CO2
Therefore, 26.48 grams of CH4N2O can be produced, and 0.080 grams of CO2 will remain in excess at the end of the reaction.