a.
2NH3 + 3CuO ==> N2 + 3Cu + 3H2O
First this is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants.
b.
mols NH3 = grams/molar mass = ?
mols CuO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols NH3 to mols N2.
Do the same and convert mols CuO to mols N2.
You will get different values for mols N2 so one of those must be wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent producing that smaller value is the LR.
c.
Obviously the OTHER reactant is the excess reagent.
d.
grams = mols x molar mass = ? This is the theoretical yield (TY).
e.
The actual yield (AY) is in the problem at 10.0 grams.
%yield = (AY/TY)*100 = ?
When ammonia was reacted to copper(ll)oxide at a high temperature, it will produce nitrogen gas, copper metal and water vapour. an experiment was done with 18.0g of ammonia was reacted with 90.0g of copper(ll)oxide.
(a)Write balance chemical equations for this reaction.
(b)Determine the limiting reactant.
(c)Determine excess reactant
(d)Calculate mass of nitrogen gas obtain
(e)Calculate percentage yield if only 10.0g of nitrogen gas obtain.
1 answer
1 answer