I think it’s C but I’m not sure
Oxygen and carbon dioxide
<u>Answer:</u> The correct answer is 1.18 g.
<u>Explanation:</u>
We are given a chemical equation:
We know that at STP conditions:
22.4L of volume is occupied by 1 mole of a gas.
So, 2.21L of carbon dioxide is occupied by = 
of carbon dioxide gas.
By Stoichiometry of the above reaction:
1 mole of carbon dioxide gas is produced by 1 mole of carbon
So, 0.0986 moles of carbon dioxide is produced by = 
of carbon.
Now, to calculate the mass of carbon, we use the equation:
Moles of carbon = 0.0986 mol
Molar mass of carbon = 12 g/mol
Putting values in above equation, we get:
Hence, the correct answer is 1.18 g.
Answer:
Percent yield = 90.5%
Explanation:
Given data:
Mass of carbon dioxide = 500 g
Mass of water = excess
Actual yield of carbonic acid = 640 g
Percent yield = ?
Solution:
Balanced chemical equation:
CO₂ + H₂O → H₂CO₃
Number of moles of carbon dioxide
Number of moles = Mass / molar mass
Number of moles = 500 g/ 44 g/mol
Number of moles = 11.4 mol
Now we will compare the moles of H₂CO₃ with CO₂.
CO₂ : H₂CO₃
1 : 1
11.4 : 11.4
Mass of carbonic acid:
Mass = number of moles × molar mass
Mass = 11.4 mol × 62.03 g/mol
Mass = 707.14 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 640 g/ 707.14 g × 100
Percent yield = 90.5%
The mass of Hydrogen is 2 g/mol
The mass of Helium is 4 g/mol
