General Chemistry 1
What we offer.
We provide hundreds of practice problems that focus on subject material and the all-important “why” in our General Chemistry 1 Course. We deliver learning in a concise, reliable, and easy to understand manner, giving students meaningful focus in their studies.
Below are sample problems with worked solutions within the course.
Pb(NO3)2(aq) + NaI(aq) → PbI2(s) + NaNO3(aq)
We first need to ensure that the equation is balanced to establish proper stoichiometry. Depending on your studies,
this question may not be asked so always make sure the chemical equation is properly balanced!
Here, the equation is unbalanced and can be fixed by adding 2 NaI and 2 Na(NO3)2 on the reactants
and products side, respectively. Giving us the following chemical equation:
Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)
Now we can determine which of the two is the limiting reagent. We can see that there are 2 moles of NaI needed for
every one mole of lead(II) nitrate. This measn we will need at least 2 times the molar amount for it to not be the
limiting reagent.
Calculating the moles of lead(II) nitrate present first, using its molar mass:
16.18 g Pb(NO3)2 × 1 mol 331.20 g = 0.049 mol of Pb(NO3)2
Now, we can calculate the amount of NaI used, in moles, with the same method. As always, what you are converting to is on top!
12.00 g NaI × 1 mol 149.89 g = 0.080 mol of NaI
At first glance, we can see that there are not twice as many moles of NaI as lead(II) nitrate. This means that
NaI is the limiting reagent in the reaction.
All of the NaI will be chewed up and used in the reaction, which means we now need to find the maximum amount of moles of
PbI2 that can be generated by using our chemical equation and the 2:1 ratio.
We can set up the conversion, keeping in mind what we are converting to is on top. We can vocally say "One mole of
PbI2 is made per two moles of NaI":
0.080 mol NaI × 1 mol PbI2 2 mol NaI = 0.040 mol PbI2
To find the number of grams of PbI2 made, we can now simply use its molar mass and multiply accordingly:
0.040 mol PbI2 × 461.01 g 1 mol = 18.45 g PbI2
A total of ~18.45 g of lead(II) iodide are made from this reaction. Take note that by mass, we have more solid of our precipitate that we did solid of our two other compounds. This is expected when two very heavy atoms combine to make a new compound, and emphasizes why it is important to go through moles!
Some helpful conversions:
1 L = 0.264 gal
Density of Gasoline = 0.749 g/mL
Though this problem does not have anything specific to do with chemistry, it is important to understand the logic and critical thinking of figuring out these sor tof complex dimensional analysis problems.
The first step in solving problems with this sort of word complexity is to boil it down to exactly what the problem is looking for.
Here, it is asking for gallons of gas. We should begin by looking at what values have volume in it: density of gasoline.
Next, the fuel flow limit is given in mass. Therefore we need to look at what units have mass in it and it is again the density.
So from the mass in a two hour time period (the period of the race) we can convert to volume using the density then ultimately to gallons using the conversion of liters to gallons.
The logic of solving the problem, by using unit conversions, can be illustrated below:
Fuel Flow Limit → Mass of Fuel Needed in Two Hours → Mass to Liters → Liters to Gallons
To begin the conversion, we can start with the fact that there are two hours per race (we are converting to "mass of fuel per race" here, so hours will cancel and mass will still be in the numerator):
100 kg/hour × 2 hours 1 race = 200 kg/race
Now, we can convert this to grams so we can calculate the volume of fuel using the density. Because we are wanting to convert to mL, we will divide the mass of fuel by the density (so the volume will be on top, in the numerator). Note that the kg have been converted to grams by multiplying by 1000:
200,000 g/race × 1 mL 0.749 g = 267,000 mL/race
We can convert this to liters by dividing by 1000, giving us:
267 L/race
We can now convert this to gallons. Knowing that there is 0.264 gal per liter, we can multiply our value of liters by 0.264, giving us:
267 L/race × 0.264 gal 1 L = 70.5 gal/race
The car would need approximately 71 gallons to complete the 2 hour race!
A. n = 3, ℓ = 2, mℓ = +2, ms = -1/2
B. n = 3, ℓ = 1, mℓ = −1, ms = +1/2
C. n = 3, ℓ = 2, mℓ = +3, ms = −1/2
D. n = 3, ℓ = 3, mℓ = 0, ms = −1/2
The correct answer is choice A. Because ℓ = 2, which corresponds to a d-orbital. The value of mℓ = +2 is valid,
and the "spin down" ms is -½.
Options B and D are incorrect because they correspond to a p-orbital and f-orbital, respectively, because of their values of ℓ.
Choice C is incorrect because the mℓ value is greater than the value of ℓ, which is not allowed.