Question

A 25.5 liter balloon holding 3.5 moles of carbon dioxide leaks. How many grams of carbon dioxide escaped before the container could be sealed if the new volume of the balloon is 15.4L?

Answer 1

Assumptions made: 
1. Temperature and pressure is constant. 
2. Ideal gas is involved. 

The variables involved in this problem is the moles of the gas and the volume. We find their relationship from the ideal gas law. PV=nRT. With all other concept besides n and V is constant, Volume is directly proportional to the amount of gas inside the balloon. Solving this is as simple as ratio and proportion. 

25.5L/3.5 moles = =15.4L/X moles 
X= 2.11 moles. 

Answer 2

Answer:

61 grams

Explanation:

If the number of moles of CO₂ in the balloon is given to be 3.5 moles, the mass of CO₂ in the balloon will be

number of moles = mass ÷ molar mass

Hence, mass = number of moles × molar mass

Molar mass of CO₂ is 44g. This is 44 because the atomic mass of carbon is 12 while that of oxygen is 16. Thus, 12 + (16 × 2) = 44 g

mass =  3.5 moles × 44g

mass = 154g (which is the initial mass)

When the mass of CO₂ in the 25.5 liter balloon is 154g, the mass of CO₂ in the balloon when the volume of CO₂ in the balloon is 15.4 liter will be X

To get X,

25.5 L ⇒ 154g

15.4 L ⇒ X

cross-multiply, and

X = (15.4 × 154) ÷ 25.5

X = 93.00 grams (which is the final mass)

93.00 grams of CO₂ was left in the balloon, hence the mass of CO₂ that escaped will be: initial mass minus final mass

= 154g - 93g

= 61g

The mass of CO₂ that escaped is 61 grams