Answer:
13mL of CO₂ are produced
Step-by-step explanation:
Based on the chemical reaction:
2CO(g) + O₂(g) → 2CO₂(g)
Where 1 mole of O₂ in excess of CO produce 2 moles of CO₂}
Thus, to solve this problem we need to convert the mass of oxygen gas to moles and, using the chemical reaction we can find the moles of CO₂ and its volume at STP:
Moles O₂ -Molar mass: 32g/mol-
55g O₂ * (1mol / 32g) = 1.72 moles O₂
Moles CO₂:
1.72 moles O₂ * (2 moles CO₂ / 1mol O₂) = 3.44 moles CO₂
Volume:
PV = nRT
V = P / nRT
Where P is pressure = 1atm at STP
n are moles = 3.44 moles of CO₂
R is gas constant = 0.082atmL/molK
T is absolute temperature = 273.15K at STP
V = 1atm / 3.44mol*0.082atmL/molK*273.15K
V = 0.0130L = 13mL of CO₂ are produced