Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.
Solution:- The balanced equation for the combustion of acetylene is:

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

= 
The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.
From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

= 
Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.
So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

= 
Where is the picture to this?
Answer:
just use the tongs and put it on a plate
Explanation:
The pressure in atm exerted by 1 mole of methane placed into a bulb with a volume of 244.6 mL at 25°C is 101.94atm.
<h3>How to calculate pressure?</h3>
The pressure of an ideal gas can be calculated using the following formula:
PV = nRT
Where;
- P = pressure
- V = volume
- n = number of moles
- R = gas law constant
- T = temperature
According to information in this question;
- T = 25°C = 25 + 273 = 298K
- V = 244.6mL = 0.24L
- R = 0.0821 Latm/Kmol
P × 0.24 = 1 × 0.0821 × 298
0.24P = 24.47
P = 24.47/0.24
P = 101.94atm
Therefore, the pressure in atm exerted by 1 mole of methane placed into a bulb with a volume of 244.6 mL at 25°C is 101.94atm.
Learn more about pressure at: brainly.com/question/11464844