Lab 10: Molar Volume of a Gas & the Percent KClO3 in a Mixture
|Only help on the Problems in the lab manual are available at this time.|
PROBLEMS (page 98)
1. 2.037 g of a mixture of NaClO3 and NaCl (inert) was heated until all of the NaClO3 has decomposed. The mass of the residue was 1.610 g. Calculate the % NaClO3 in the mixture.
__ NaClO3 (s) --> __NaCl (s) + __O2(g)
NaCl(s) --> NaCl(s)
This problem is very similar to the problems in Lab 4. That's where hydrated salts were heated and the water was driven off. It also had NaCl as an inert component of the mixture. The chemical equation is given so that you can balance it. That's needed to know the ratio of the decomposition products compared to the original NaClO3. In the first equation, we can see that Na and Cl are already balanced. However, there are 3 oxygen atoms on the left side and only 2 on the right side. So that will require us to double NaClO3 and triple O2. That will make the oxygen atoms balance.
2NaClO3 (s) --> 2NaCl (s) + 3O2 (g)
The approach to solving the problem is to realize that the grams lost by the mixture is from the loss of oxygen gas. So we had 2.037 g and ended with 1.610 g. That's a loss of 0.427 grams of oxygen gas. That can be changed to moles of oxygen gas. Then using the balanced equation, we can find the moles of NaClO3 that must have decomposed. The moles of NaClO3 is then converted to grams of NaClO3. That will be a fraction of the original mixture, which leads to our percent by weight. The question didn't specify % by weight, but we will assume that's what is wanted.
To find percent, they say to multiply by 100, but you are actually multiplying by 100/100. You just don't divide by 100 because you see "/100". You simply turn that "/100" into the % sign. "%" means "per 100", so that means there is a 100 still in the denominator.
|2. In the above experiment (problem 1) a student collected 343 mL of gas by water displacement at 23 °C and 731 torr. Calculate the molar volume of O2 at STP from these results. Also calculate the percent error in the student's value.
First we assume that the 343 mL of gas is O2, coming from the decomposition of NaClO3. If this 343 mL were at 0°C and 760 torr, then at standard conditions, it would still be 343 mL. However, 23°C is warmer than standard conditions and 731 torr is less pressure than standard conditions. So you need to decide if the changing to STP conditions will cause the volume to go up or down. If temperature goes from 23°C to 0°C, then the gas molecules are moving slower and they bang against with less force causing them to shrink more (less volume). To find out by how much we have to turn Celsius temperatures into Kelvin to get them on the proper scale. 0°C is 273K and 23°C is 296K. Now we see their perspective sizes. To make the volume go down, it makes sense to use the fraction of 273K/296K.
Going from 731 torr to 760 torr means the gas will feel more pressure, causing the molecules to squeeze together more (less volume). The exact amount depends on a fraction based on the starting and final pressures. So a fraction of 731 torr/760 torr, will reduce the volume.
Molar volume is "Liters per mole". So we need to take the volume in liters and divide by moles. The volume of 343 mL adjusted to STP will find the volume (Liters). Problem 1 gives us the grams of O2. So we start with finding liters per gram, but then change the grams of O2 to moles. See rows 5 to 7.
The final answer is in the ballpark of the expected 22.4 liters per mole that all ideal gases are suppose to have. To get percent error this value, subtract 22.4 from cell H6 and then divide by 22.4. Then multiply by 100
3. How much KClO3 must be decomposed to provide 400. mL of oxygen gas collected at STP?
__KClO3 (s) --> __KCl (s) + __O2 (g)
The balancing of this equation is just like the one in Problem 1. Instead of sodium chlorate (NaClO3, we have potassium chlorate (KClO3).
2KClO3 (s) --> 2KCl (s) + 3O2 (g)
Practice Problem for #3. These calculations are useful in survival types of scenarios. For example, let's say you supervise a mining operation and want the workers to carry around an oxygen source of KClO3 that could provide 10 minutes of oxygen. A person needs about 140 milliliters of oxygen per minute, so that is 14000 mL of oxygen for the 10 minutes. How many grams of KClO3 must be decomposed to provide 14000 mL of oxygen gas at STP? Also, what would be mL volume of the KClO3 powder?
|4. The volume of a sample of gas that weighs 0.324 g is 295 mL at 26°C and 732 torr. Calculate the density of the gas at STP.
Density of gases is normally written in grams per liter (grams divided by liters. You are given the grams and the millilters, so the current density is 0.324 g/295mL = 1.098 g/L. However, it wants the density at STP. So there is an adjustment of the volume when going from 26°C to 0°C and 732 torr to 760 torr. Like in Problem 2, you decide if these changes raise or reduce the volume (mass doesn't change). That guides you on how to set up the fraction. Also, like before, temperature's always needs to be converted to the Kelvin scale first. So 0°C is 273K and 26°C is 299K.
|5. What is the molar mass of a gas if 0.528 g of it occupies 457 mL when collected at STP.
This problem is easier than Problem 4. There are no adjustments due to temperature or pressure. So it's just rows 5, 6 and 7 in the above spreadsheet plus the multiplication of 22.4 liters per mole as mentioned above. Hint: The molar mass is close to the gas, acetylene, which is used in acetylene torches for welding.