Answer:
6.05g
Explanation:
The reaction is given as;
Ethane + oxygen --> Carbon dioxide + water
2C2H6 + 7O2 --> 4CO2 + 6H2O
From the reaction above;
2 mol of ethane reacts with 7 mol of oxygen.
To proceed, we have to obtain the limiting reagent,
2,71g of ethane;
Number of moles = Mass / molar mass = 2.71 / 30 = 0.0903 mol
3.8g of oxygen;
Number of moles = Mass / molar mass = 3.8 / 16 = 0.2375 mol
If 0.0903 moles of ethane was used, it would require;
2 = 7
0.0903 = x
x = 0.31605 mol of oxygen needed
This means that oxygen is our limiting reagent.
From the reaction,
7 mol of oxygen yields 4 mol of carbon dioxide
0.2375 yields x?
7 = 4
0.2375 = x
x = 0.1357
Mass = Number of moles * Molar mass = 0.1357 * 44 = 6.05g
Answer:
22.656 grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C
Explanation:
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law:
P * V = n * R * T
where R is the molar constant of the gases and n the number of moles.
In this case you know:
- R= 0.082
- T= 24 °C= 297 °K (being 0°C=273°K)
Replacing:
Solving:
n=0.708 moles
Knowing that oxygen gas is a diatomic gas of molecular form O₂ and its mass is 32 g / mole, you can apply the following rule of three: if 1 mole contains 32 grams, 0.708 moles, how much mass will it have?
mass= 22.656 grams
<u><em>22.656 grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C</em></u>
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