3PbO2 + 4AlCl3 — 3PbCl2 + 2A1203 Which conversion factor will you need to calculate the mass
of PbC14 formed when 17 grams of AlCl3 react?

Explanation:

2 Al + 3 Cl2 → 2 AlCl3

Suppose we know we have 123.2 g of Cl2. How can we determine how many moles of AlCl3 we will get when the reaction is complete? First and foremost, chemical equations are not balanced in terms of grams; they are balanced in terms of moles. So to use the balanced chemical equation to relate an amount of Cl2 to an amount of AlCl3, we need to convert the given amount of Cl2 into moles. We know how to do this by simply using the molar mass of Cl2 as a conversion factor. The molar mass of Cl2 (which we get from the atomic mass of Cl from the periodic table) is 70.90 g/mol. We must invert this fraction so that the units cancel properly:

Screen Shot 2014-07-21 at 5.16.11 PM

Now that we have the quantity in moles, we can use the balanced chemical equation to construct a conversion factor that relates the number of moles of Cl2 to the number of moles of AlCl3. The numbers in the conversion factor come from the coefficients in the balanced chemical equation:

Screen Shot 2014-07-21 at 5.16.18 PM

Using this conversion factor with the molar quantity we calculated above, we get

Screen Shot 2014-07-21 at 5.16.21 PM

So, we will get 1.159 mol of AlCl3 if we react 123.2 g of Cl2.

In this last example, we did the calculation in two steps. However, it is mathematically equivalent to perform the two calculations sequentially on one line:

Screen Shot 2014-07-21 at 5.16.27 PM

The units still cancel appropriately, and we get the same numerical answer in the end. Sometimes the answer may be slightly different from doing it one step at a time because of rounding of the intermediate answers, but the final answers should be effectively the same.

Example 10

How many moles of HCl will be produced when 249 g of AlCl3 are reacted according to this chemical equation?

2 AlCl3 + 3 H2O(ℓ) → Al2O3 + 6 HCl(g)

Solution

We will do this in two steps: convert the mass of AlCl3 to moles and then use the balanced chemical equation to find the number of moles of HCl formed. The molar mass of AlCl3 is 133.33 g/mol, which we have to invert to get the appropriate conversion factor:

187mol

Now we can use this quantity to determine the number of moles of HCl that will form. From the balanced chemical equation, we construct a conversion factor between the number of moles of AlCl3 and the number of moles of HCl:

6molhcl

Applying this conversion factor to the quantity of AlCl3, we get

187mol

Alternatively, we could have done this in one line:

249galcl3

The last digit in our final answer is slightly different because of rounding differences, but the answer is es

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890