Answer:
22.9 Liters CO(g) needed
Explanation:
2CO(g) + O₂(g) => 2CO₂(g)
? Liters 32.65g
= 32.65g/32g/mol
= 1.02 moles O₂
Rxn ratio for CO to O₂ = 2 mole CO(g) to 1 mole O₂(g)
∴moles CO(g) needed = 2 x 1.02 moles CO(g) = 2.04 moles CO(g)
Conditions of standard equation* is STP (0°C & 1atm) => 1 mole any gas occupies 22.4 Liters.
∴Volume of CO(g) = 1.02mole x 22.4Liters/mole = 22.9 Liters CO(g) needed
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*Standard Equation => molecular rxn balanced to smallest whole number ratio coefficients is assumed to be at STP conditions (0°C & 1atm).
The formula for kinetic energy is KE=1/2(mv²). Since both mass and velocity are multiplied by each other, particle with a larger mass needs to be moving slower than a particle with less mass if both have the same kinetic energy. You can think of it as 2KE/m=v² or 2KE/v²=m, If you increase the mass the velocity needs to decrease to keep the same KE value.
I hope this helps. Let me know in the comments if anything is unclear.
