How many moles of gas does it take to occupy 120 L at a pressure of 233 kPa and a temperature of 340 K?
2 answers:
Answer:
The correct answer is option D.
Explanation:
Using ideal gas equation:
PV = nRT
where,
P = Pressure of gas = 233 kPa
1 atm = 101.325 kPa
V = Volume of gas = 120 L
n = number of moles of gas = ?
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of gas = 340 K
Putting values in above equation, we get:
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Taking into account the reaction stoichiometry, 2.13 grams of magnesium was dissolved in the solution.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
Mg²⁺(aq) + Na₂CO₃(aq) → MgCO₃(s) + 2 Na⁺(aq)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Mg²⁺: 1 mole
- Na₂CO₃: 1 mole
- MgCO₃: 1 mole
- Na⁺: 2 moles
The molar mass of the compounds is:
- Mg²⁺: 24.3 g/mole
- Na₂CO₃: 106 g/mole
- MgCO₃: 84.3 g/mole
- Na⁺: 23 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Mg²⁺: 1 mole ×24.3 g/mole= 24.3 grams
- Na₂CO₃: 1 mole ×106 g/mole= 106 grams
- MgCO₃: 1 mole ×84.3 g/mole=84.3 grams
- Na⁺: 2 moles ×23 g/mole= 46 grams
The following rule of three can be applied: If by reaction stoichiometry 1 mole of Na₂CO₃ react with 24.3 grams of magnesium, 0.0877 moles of Na₂CO₃ react with how much mass of magnesium?
<u><em>mass of magnesium= 2.13 grams</em></u>
Finally, 2.13 grams of magnesium was dissolved in the solution.
Learn more about the reaction stoichiometry: