For Iron:
For Oxygen:
These are the two chemical symbols for the two elements found in Iron Oxide.
Answer:
0.162 moles of CO₂ are produced by this reaction
Explanation:
The reaction is:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) +4H₂O(g)
As we have the volume of propane, we need to know the mass that has reacted, so we apply density's concept.
Density = Mass / Volume → Density . Volume = Mass
0.00183 g/mL . 1300 mL = Mass → 2.379 g
We determine the moles → 2.379 g . 1mol / 44 g = 0.054 moles
Ratio is 1:3. 1 mol of propane can produce 3 moles of dioxide
Then, 0.054 moles may produce (0.054 .3)/1 = 0.162 moles
The partial pressure of methane in the mixture of methane and ethane has been 1 atm.
Partial pressure has been the pressure exerted by a gas in the solution or mixture. The partial pressure of each gas has been the total pressure of the gaseous mixture.
The partial pressure of the gas has been dependent on the volume, temperature, and concentration of the gas.
The given methane has a partial pressure of 1 atm in the 15 L vessel. The addition of ethane results in the change in the total pressure of the mixture, as there have been additional moles of solute that contributes to the solution pressure.
However, since there has been no change in the concentration and volume of methane, the pressure exerted by methane has been the same. Thus, the partial pressure of methane has been 1 atm.
For more information about the partial pressure, refer to the link:
brainly.com/question/14623719
Answer:
Db
Dubnium/Symbol
Explanation:
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Answer:
Mass = 8.46 g
Explanation:
Given data:
Mass of water produced = ?
Mass of glucose = 20 g
Mass of oxygen = 15 g
Solution:
Chemical equation:
C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂
Number of moles of glucose:
Number of moles = mass/molar mass
Number of moles = 20 g/ 180.16 g/mol
Number of moles = 0.11 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 15 g/ 32 g/mol
Number of moles = 0.47 mol
now we will compare the moles of water with oxygen and glucose.
C₆H₁₂O₆ : H₂O
1 : 6
0.11 : 6/1×0.11 = 0.66
O₂ : H₂O
6 : 6
0.47 : 0.47
Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 0.47 mol ×18 g/mol
Mass = 8.46 g
