Help pleasee confused on these questions.

Question

3. 2Al+ 3Cl 2 → 2AlCl 3

How many grams of AlCl 3 could be produced from 34.0 g Al and 39.0 g

Cl 2 ?

4. N 2 + 3H 2 → 2NH 3

a. How many grams of NH 3 will be produced if you have 55.2 g N 2 and

89.7 g of H 2 ?

b. What are your limiting reagents?

5. C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O

a. If you have 17.6 g of C 3 H 8 and 36.2 g of O 2 , how much CO 2 will be

Made? 

Answer 1

3. Based on the following reaction, how many grams of AlCl₃ could be produced from 34.0 g Al and 39.0 g Cl₂?

2Al + 3Cl₂ → 2AlCl₃

This is a limiting reagent problem. We need to determine which reactant will produce less product, based on how many moles of each reactant we start with. We can only produce as much product as the limiting reactant allows. Keep in mind that we must use the molar masses (g/mol) of each compound as well as the balanced reaction stoichiometry to solve this problem.

First thing to do is to calculate the molar masses of our compounds, using the periodic table:

MM_Al = 26.98 g/mol

MM_Cl2 = 70.90 g/mol

MM_AlCl3 = 133.33 g/mol

Next, we calculate the maximum moles of product we can form from each reactant:

Moles of AlCl₃ from 34.0g of Al:

34.0 g Al ( 1 mol Al ) = 1.26 mol Al ( 2 mol AlCl₃) = 1.26 mol AlCl₃

( 26.98 g Al ) ( 2 mol Al )

Moles of AlCl₃ from 39.0g of Cl₂:

39.0 g Cl₂ ( 1 mol Cl₂ ) = 0.55 mol Cl₂ ( 2 mol AlCl₃) = 0.37 mol AlCl₃

( 70.9 g Cl₂ ) ( 3 mol Al )

Based on our calculations, it's apparent that 39.0 g of Cl₂ will only allow us to produce 0.37 mol of AlCl₃, whereas 37.0 g of Al will allow us to produce much more product (1.26 mol). So, Cl₂ is our limiting reagent, and 0.37 mol of AlCl₃ is the theoretical max. yield for this reaction.

The question asks for the theoretical product in grams, not moles, so we must do one final conversion:

Theoretical Yield of AlCl₃ in moles:

0.37 mol AlCl₃ ( 133.33 g AlCl₃ ) = 49.33 g AlCl₃

( 1 mol AlCl₃ )

4. This follows the same format as Q3.

5. This follows the same format as Q3.