Considering the ideal gas law, the volume of gas produced at 25.0 °C and 1.50 atm is 184.899 L.
<h3>Definition of ideal gas</h3>
An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
<h3>Ideal gas law</h3>
An ideal gas is characterized by absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of gases:
P×V = n×R×T
<h3>Volume of gas</h3>
In this case, you know:
- P= 1.50 atm
- V= ?
- n= 500 g×= 11.36 moles, being 44 the molar mass of CO₂
- R= 0.082
- T= 25 C= 298 K (being 0 C=273 K)
Replacing in the ideal gas law:
1.50 atm×V = 11.36 moles×0.082 × 298 K
Solving:
V= (11.36 moles×0.082 × 298 K) ÷ 1.50 atm
<u><em>V= 184.899 L</em></u>
Finally, the volume of gas produced at 25.0 °C and 1.50 atm is 184.899 L.
Learn more about the ideal gas law:
<u>brainly.com/question/4147359?referrer=searchResults</u>
The volume of oxygen required to burn 12.00 L ethane is calculated as follows
find the moles of C2H6 used
At STP 1 mole is always = 22.4 L, what about 12.00 L
= ( 12.00L x 1 moles) 22.4 L = 0.536 moles
write the reacting equation
2C2H6+ 7O2 = 4CO2 + 6H2O
by use of mole ratio between C2H6 :O2 which is 2:7 the moles of O2
= 0.536 x7/2= 1.876 moles
again at STP 1mole = 22.4 L what about 1.876 moles
= 22.4 L x 1.876 moles/ 1 mole = 42.02 L
Answer:
Just add what they equal to hope this helps :)
Explanation: