What volume is occupied by 0.109 molmol of helium gas at a pressure of 0.98 atmatm and a temperature of 307 K
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Answer:
10.3125 grams of carbon dioxide will be formed.
Explanation:
The balanced reaction is:
C₃H₈ + 5 O₂→ 3 CO₂ + 4 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles participate in the reaction:
- C₃H₈: 1 mole
- O₂: 5 moles
- CO₂: 3 moles
- H₂O: 4 moles
Being the molar mass of the compounds:
- C₃H₈: 44 g/mole
- O₂: 32 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
then by stoichiometry of the reaction, the following amounts of mass participate in the reaction:
- C₃H₈: 1 mole* 44 g/mole = 44 g
- O₂: 5 moles* 32 g/mole= 160 g
- CO₂: 3 moles* 44 g/mole= 132 g
- H₂O: 4 moles* 18 g/mole= 72 g
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, it is possible to use a simple rule of three as follows: If by reaction stoichiometry 44 grams of propane react with 160 grams of oxygen, 7.2 grams of propane react with how much mass of oxygen?
mass of oxygen= 26.18 grams
But 26.18 moles of O₂ are not available, 12.5 grams are available. Since you have less mass than you need to react with 7.2 grams of propane, oxygen O₂ will be the limiting reagent.
Then you can apply the following rule of three: if by stoichiometry of the reaction 160 grams of oxygen form 132 grams of carbon dioxide, 12.5 grams of oxygen will form how much mass of carbon dioxide?
mass of carbon dioxide= 10.3125 grams
<u><em>10.3125 grams of carbon dioxide will be formed.</em></u>
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