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.
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
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