Activity A (continued from previous page) 4. Explore: Click Reset and change the indicator to Bromthymol blue. Add exactly 8.8 mL of HNO3 to the flask. A. What does the color of the indicator tell you about the current pH of the flask? B. Add one more drop of HNO3. What does the color tell you about the pH now? ¬¬¬¬¬¬¬ C. If you combine the results of this question with the results from question 3B, what do you know about the total pH change caused by adding the last 0.1 mL of HNO3? 5. Apply: Water has a pH of 7. If 0.1 mL (about one drop) of 1.0 M HNO3 is added to 100 mL of water, the result is a solution with a concentration of 0.001 M HNO3. A. What is the pH of 0.001 M HNO3? B. How much did one drop of HNO3 cause the pH of water to change? C. How does this relate to what you determined in question 4C?

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

explanation:

a