Consider the following reaction: iron (s) + chlorine (g) à iron (iii) chloride a. write the balanced chemical equation. b. 25.0 g of iron reacts with excess chlorine gas. a. calculate the moles of iron reactant. b. calculate the moles of iron (iii) chloride. c. calculate the mass of iron (iii) chloride.

Explanation:

                  Fe + Cl₂ →  FeCl₃

a. The balanced equation

    All chemical equations obey the law of conservation of matter. To balance this above equation, we either inspect or use mathematical method to obtain a balanced equation.

We put coefficients a, b and c at the back of the compounds as shown below:

                    aFe + bCl₂ →  cFeCl₃

For Fe:

      a = c                          (i)

For Cl:

     2b = 3c                        (ii)

let a = 1, c= 1

Solving for the unknown b, we have:

          b =

           b =

The complete reaction equation is therefore:

            Fe + Cl₂ →  FeCl₃

                                     or

                       2Fe + 3Cl₂ →  2FeCl₃

Problem a:

Mass of reacting iron = 25g

unknown: number of moles of iron

Solution

To find the number of moles, we apply the mole concept using the expression below:

To find the molar mass of the reactant iron, we know atomic mass of iron is 56g.

The molar mass is therefore, 56gmol⁻¹

number of moles of the iron reactant  = 0.45mole

Problem b:

From the balanced equation of the reaction:

             2Fe + 3Cl₂ →  2FeCl₃

      2 moles of Fe produces 2 moles of FeCl₃

     So 0.45mole of Fe would produce 0.45 mole of FeCl₃

Problem C:

Applying the mole concept;

      mass of FeCl₃ = number of moles of FeCl₃ x molar mass of FeCl₃

     number of moles of FeCl₃ = 0.45mole

     molar mass of FeCl₃ = ?

Atomic mass of Fe = 56g

                          Cl = 35.5g

      Molar mass of FeCl₃ = 56 + (3 x 35.5) = 56 + 106.5 = 162.5gmol⁻¹

mass of FeCl₃ = 0.45 x 162.5 = 73.1g

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%