**Chemistry- Gas Law Problems, Please Help?**

1. Not enough air is pumped into an inflatable life raft to make it completely fill out. The atmospheric pressure is 14.7 psi. What is the pressure of the air inside the raft? Explain.

2. How many liters of sulfur dioxide gas will be needed to react completely with 3.85 L of oxygen gas to form sulfur trioxide?

3. How many liters of oxygen gas are needed for the combustion of 2.05 L of ethane gas?

1. Using ideal gas law, P=nRT/V, weight the liferaft, ML (without air), weight the liferaft after filled with air, Mtot, find the air weigh inside the liferaft, Mair = Mtot – ML.

Check the volume of air needed to fill the liferaft.

molecular weight of air = 29, hence mol, n = Mair / 29

Therefore, Pressure = Mol of Air x Gas Constant, R x Temperature of Air / Volume of Air

2. Chemical Reaction: 2 SO2 + O2 —> 2 SO3

1 mol of O2 = 32g, 1 mol of gas at STP occupied 22.4L, density of gas at STP = 0.04464mol/L

Hence, 3.85L O2 = 0.1718 mol O2

1 mol of O2 need 2 mol of SO2 to react completely, hence 0.1718mol O2, need 0.344 mol SO2

Same as above, 1 mol of gas at STP = 0.04464 mol/L

Hence, you need 0.344/0.04464 = 7.706 L of SO2 for the complete reaction

3.Chemical Reaction: C2H6 + 3.5O2 –> 2CO2 + 3H2O

1 mol of Ethane need 3.5 mol of Oxygen for complete combustion

density of gas at STP = 1/22.4 mol/L = 0.04464 mol/L

2.05 L of Ethane contained 0.04464 x 2.05 = 0.0915 mol

Hence, 0.0915 mol of ehtane need 0.3203 mol of O2

0.3203 mol of O2 = 0.3203/0.04464 = 7.175 L O2

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