Answer:
Percent yield = 57%
Explanation:
Given data:
Number of moles of propane = 4.50 mol
Number of moles of carbon dioxide = 7.64 mol
Percent yield = ?
Solution:
Chemical equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Now we will compare the moles of propane and carbon dioxide.
C₃H₈ : CO₂
1 ; 3
4.50 : 3×4.50 = 13.5 mol
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 7.64 mol / 13.5 mol × 100
Percent yield = 0.57× 100
Percent yield = 57%
Arrow on the table below draw an
Moles of MgS2O3 = 223/molar mass of MgS2O3
= 223/136.42
= 1.634 moles.
Hope this helps!
Here is the full question:
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.
What is the concentration at 10 minutes? (Round your answer to three decimal places.
Answer:
0.046 %
Explanation:
The rate-in;

= 0.8
The rate-out
= 
= 
We can say that:

where;
A(0)= 0.2% × 6000
A(0)= 0.002 × 6000
A(0)= 12

Integration of the above linear equation =

so we have:



∴ 
Since A(0) = 12
Then;



Hence;



∴ the concentration at 10 minutes is ;
=
%
= 0.0456667 %
= 0.046% to three decimal places