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Answer:
The volume will be 568.89 mL.
Explanation:
Boyle's law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure"
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
or P * V = k
Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. That is, the pressure of the gas is directly proportional to its temperature. Gay-Lussac's law can be expressed mathematically as follows:
Where P = pressure, T = temperature, K = Constant
Finally, Charles's law indicates that as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. In summary, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:
Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:
Studying an initial state 1 and a final state 2, it is fulfilled:
In this case:
- P1= 960 mmHg
- V1= 550 mL
- T1= 200 C= 473 K (being 0 C=273 K)
- P2= 830 mmHg
- V2= ?
- T2= 150 C= 423 K
Replacing:
Solving:
V2= 568.9 mL
<u><em>The volume will be 568.89 mL.</em></u>
<u>Answer:</u> The limiting reagent is oxygen gas.
<u>Explanation:</u>
Limiting reagent is defined as the reactant that is present in less amount and it limits the formation of products.
Excess reagent is defined as the reactant which is present in large amount.
To calculate the number of moles, we use the equation:
.....(1)
- <u>For propane:</u>
Given mass of propane = 30.0 g
Molar mass of propane = 44.1 g/mol
Putting values in equation 1, we get:
- <u>For oxygen:</u>
Given mass of oxygen = 75.0 g
Molar mass of oxygen = 32 g/mol
Putting values in equation 1, we get:
The chemical equation for the combustion of propane follows:
By Stoichiometry of the reaction:
5 moles of oxygen gas reacts with 1 mole of propane.
So, 2.34 moles of oxygen gas will react with = of propane
As, given amount of propane is more than the required amount. So, it is considered as an excess reagent.
Thus, oxygen is considered as a limiting reagent because it limits the formation of product.
Hence, the limiting reagent is oxygen gas.