AP Chemistry - Stoichiometry

You may have heard the famous story of Archimedes and the golden crown; where the ancient Greek mathematician was challenged to determine if his King's golden crown was truly made of 100% gold. In that story Archimedes used density to determine that the crown was not 100% gold.

The density of gold is $19.3\frac{g}{cm^3}$

In real life there are many applications for this concept that you can use in real life. For example lets say that you are in the business of selling scrap metal. You have spent a lot of time collecting a yellowish metal; but have no clue what it is, and don't wish to simply trust the judgment of the scrap metal buyer. You do know that the most likely yellow metals; given your sources for the metal are copper, brass and gold. You just don't know which it is by looking at it.

The density of copper is $8.9\frac{g}{cm^3}$

The density of brass is $8.4\frac{g}{cm^3}-8.73\frac{g}{cm^3}$ the density of brass is a range because it is an alloy of copper and zinc, and can be made with different proportions of each element, whereas the density of copper is exact because it is a pure element, whose composition doesn't change (assuming none of the copper is oxidized, in which case it turns green like the statue of liberty and is no longer a yellow metal).

You have a scale and know the weight of your collected metal to be 20lbs. You take a container with water filling it to an initial volume of 40gal

Then you place your metal in the container, and the volume rises to 61.3gal. What is the identity of your metal?

1gal = 3.785L

1lb = 453.592g

Copper

Brass

Gold

None of these

Steel

Explanation

We know the following relationships from the information in the problem and from the definitions of the units:

Solving the problem:

The change in volume after you submerge your metal in the container is the volume of the metal itself. Therefore: $V_{(metal)}=61.3gal-40.0gal=21.3gal$

Since the densities we have been given are all expressed as $\frac{g}{cm^3}$ we should convert our numbers into those units as shown below:

$21.3gal*{\frac{3.785L}{1gal}}*{\frac{1000ml}{1L}}*{\frac{1cm^3}{1mL}}=80620.5cm^3=V_{(metal)}$

$20.0lb*\frac{453.592g}{1lb}=9071.8g=m_{(metal)}$

Now that we have both the mass and volume of our unknown metal you can calculate the density of the metal because density is defined as:

$density=\frac{m}{V}$

So the density of our metal is $\frac{9071.8g}{80620.5cm^3}=8.9\frac{g}{cm^3}$

Since our metal's density matches the density of copper we can now prove that the metal is pure copper.

What volume of 0.5M NaOH is necessary to neutralize 15mL of 1.0M nitrous acid?

30mL

10mL

30L

Explanation

Both the acid and the base provide one equivalent of the necessary ions (H+ and OH–, respectively) and therefore you can use the equation M1V1 = M2V2.

(1)(15) = (0.5)V2

V2 = 30mL

How many centimeters are in 14.1 inches?

Explanation

$14.1 in * \frac{2.54 cm}{1 in} = 35.8 cm$

Remember, there are 2.54 centimeters in every one inch. 2.54 can be multiplied by the given number of inches in order to cancel units, and convert the inches to centimeters. Answer with the same number of significant figures, as what was given in the initial number.

What is the mass of 25 iron atoms?

$2.32*10^{-21}g$

$2.5*10^{-22}g$

$1.23*10^{-23}g$

$2.5*10^{-21}g$

$1.85*10^{-23}g$

Explanation

First, convert the number of atoms to moles using Avogadro's number:

$25\ \text{atoms}*\frac{1\ \text{mole}}{6.022*10^{23}\ \text{atoms}}=4.1514*10^{-23}\ \text{moles Fe}$

Next, convert moles of iron to grams of iron using the molar mass:

$4.1514*10^{-23}mol\ Fe*\frac{55.85g\ Fe}{1mol\ Fe}=2.32*10^{-21}g$

How much water is necessary to make a solution with a molality of when of sodium chloride are in the solution?

Explanation

This question is asking us to solve for molality, which has the following equation:

$m=\frac{\text{moles solute}}{\text{kg solvent}}$

Note how the equation does not ask for liters of solution like molarity, but instead requires the kilograms of solvent used in the solution. By dividing the mass of the sodium chloride by its molar mass, we can solve for the moles of sodium chloride in solution.

$MM=(23)+(35.45)=58.45\frac{g}{mol}$

$moles=\frac{mass}{molar mass}=\frac{100g}{58.45\frac{g}{mol}}=1.71mol\ NaCl$

This will allow us to solve for the mass of water used in the solution.

$m=\frac{1.71mol}{x}=3.0m$

$x=\frac{1.71mol}{3.0m}=0.57kg=570g$

This gives us the mass of water required to make the solution.

The density of gold is $19.3\frac{g}{cm^3}$

The density of copper is $8.9\frac{g}{cm^3}$

You have a scale and know the weight of your collected metal to be 20lbs. You take a container with water filling it to an initial volume of 40gal

Then you place your metal in the container, and the volume rises to 61.3gal. What is the identity of your metal?

1gal = 3.785L

1lb = 453.592g

Copper

Brass

Gold

None of these

Steel

Explanation

We know the following relationships from the information in the problem and from the definitions of the units:

Solving the problem:

The change in volume after you submerge your metal in the container is the volume of the metal itself. Therefore: $V_{(metal)}=61.3gal-40.0gal=21.3gal$

Since the densities we have been given are all expressed as $\frac{g}{cm^3}$ we should convert our numbers into those units as shown below:

$21.3gal*{\frac{3.785L}{1gal}}*{\frac{1000ml}{1L}}*{\frac{1cm^3}{1mL}}=80620.5cm^3=V_{(metal)}$

$20.0lb*\frac{453.592g}{1lb}=9071.8g=m_{(metal)}$

Now that we have both the mass and volume of our unknown metal you can calculate the density of the metal because density is defined as:

$density=\frac{m}{V}$

So the density of our metal is $\frac{9071.8g}{80620.5cm^3}=8.9\frac{g}{cm^3}$

Since our metal's density matches the density of copper we can now prove that the metal is pure copper.

How many centimeters are in 14.1 inches?

Explanation

$14.1 in * \frac{2.54 cm}{1 in} = 35.8 cm$

What volume of 0.5M NaOH is necessary to neutralize 15mL of 1.0M nitrous acid?

30mL

10mL

30L

Explanation

Both the acid and the base provide one equivalent of the necessary ions (H+ and OH–, respectively) and therefore you can use the equation M1V1 = M2V2.

(1)(15) = (0.5)V2

V2 = 30mL

Consider the following chemical reaction:

$Al_2O_3(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2O(l)$

How many moles of will be produced if 75.0mL of water is produced?

1.39mol

4.17mol

1.56 moles

There is not enough information to answer this question

None of the available answers

Explanation

$75.0\hspace{1 mm}mL\times\frac{1\hspace{1 mm}g\hspace{1 mm}H_2O}{1\hspace{1 mm}mL\hspace{1 mm}H_2O}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}H_2O}{18.0\hspace{1 mm}g\hspace{1 mm}H_2O}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}Al_2(SO_4)_3}{3\hspace{1 mm}moles\hspace{1 mm}H_2O}=1.39\hspace{1 mm}moles\hspace{1 mm}Al_2(SO_4)_3$

This problem is solved based on stoichiometry. Using the molecular weights and ratios in the given reaction, we can solve for the amount of produced.

The formation of ammonia is given by the following equation:

$3H_2+N_2\rightarrow 2NH_3$

Assuming you started with 20g of hydrogen gas, how much nitrogen gas is necessary to react the hydrogen gas?

93.33g

6.67g

60g

40g

Explanation

We can convert the mass of hydrogen from grams to moles by using the equation moles = mass / molar mass

$\frac{20gH_2}{2\frac{g}{mol}H_2}=10mol H_2$

Now that we have the moles of hydrogen, we can use the molar ratio given in the equation of 1:3. Since we need 1/3 of the moles of hydrogen as nitrogen, we know we need 3.33 moles of nitrogen gas.

$10molH_2*\frac{1molN_2}{3molH_2}=3.33molN_2$

Now, we simply multiply the moles by the molar mass of nitrogen gas (28 grams per mol) and we find we need 93.33 grams of nitrogen gas.

$3.33molN_2*28\frac{g}{mol}N_2=93.33gN_2$

Stoichiometry

Help Questions

Explanation

We know the following relationships from the information in the problem and from the definitions of the units:

Solving the problem:

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We know the following relationships from the information in the problem and from the definitions of the units:

Solving the problem:

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