Question

Can some please help me complete this for me it’s due today!

Answer 1

Answer:
Mass of (NH2)2CO formed from NH3 is 10.94g
Mass of (NH2)2CO formed from CO2 is 1.94g
Percentage yield = 82.47%

Explanation:
STEP 1
For NH3
No of moles of NH3 available for reaction = Mass/Molar mass
= 6.2/17.01
=0.3645mole
From the balanced equation of reaction, we see that 2 moles of NH3 produce 1 mole of (NH2)2CO
So, 0.3645 mole of NH3 will give 0.3645/2 moles of (NH2)2CO.
No of moles of (NH2)2CO= 0.18225mole
To find the mass of (NH2)2CO that can be produced from this, we say;
Mass = Mole × Molar mass
Mass = 0.18225mol × 60g/mol
Mass of (NH2)2CO produced from NH3 is therefore 10.94g.

For CO2
No of moles = mass/molar mass
= 1.42g/44gmol-1
= 0.03227mole
From the balanced equation of reaction, we see that 1 mole of C02 gives 1 mole of (NH2)2CO.
So, 0.03227mole of CO2 will give 0.03227mole of (NH2)2CO.
Mass of (NH2)2CO from this number of mole will then be = Mole × Molar mass of (NH2)2CO
= 0.03227 × 60
= 1.94g
The amount of (NH2)2CO formed from CO2 is therefore 1.94g

STEP 2
To calculate the percentage yield, we use;
%Yield = Actual yield/Theoretical yield × 100
The actual yield has been given to be 1.6g.
To calculate the theoretical yield, we look up to the limiting reagent of the reaction as the limiting reagent determines the extent a reaction can go.
Looking at the reactants, CO2 is the limiting reagent since it has the lowest number of mole (i.e., 0.03227mole)
So, we use this number of mole to calculate the theoretical yield of (NH2)2CO
= 0.03227mol × Molar mass of (NH2)2CO (i.e., 60gmol-1)
= 1.94g

% Yield = 1.6/1.94 × 100

% Yield = 82.47%