Al2O3 (s) + 6 NaOH (aq) + 12HF (g) → 2 Na3AlF6 (s) + 9 H2O (l) In an experiment; 6.55 g Al2O3 and excess HF were dissolved in 1.75 L of 0.15 M NaOH. If 20 g Na3AlF6was obtained, a Which one is the limiting reagent? b What is the actual yield? c What is the theoretical yield? d What is the percent yield for this experiment?

A.  NaOH

B.  20 g

C.  18.4 g

D.  108%

Explanation:

Al₂O₃  +  6 NaOH  +  12 HF  ⇒  2 Na₃AlF₆  +  9 H₂O

A.  Since you have excess HF, this is not the limiting reagent.  The two possibilities are Al₂O₃ and NaOH.  Find out how many moles you have of each.  Then, use the mole ratio in the chemical equation to find out how much product each reagent can produce.  The reagent the produces the least is the limiting reagent.  The molar mass of Al₂O₃ is 101.96 g/mol.

(6.55 g Al₂O₃)/(101.96 g/mol Al₂O₃) = 0.06424 mol Al₂O₃

(0.06424 mol Al₂O₃) × (2 mol Na₃AlF₆/1 mol Al₂O₃) = 0.12848 mol Na₃AlF₆

(1.75 L NaOH) × (0.15 M NaOH) = 0.2625 mol NaOH

(0.2625 mol NaOH) × (2 mol Na₃AlF₆/6 mol NaOH) = 0.0875 mol Na₃AlF₆

NaOH produces less product, making it the limiting reagent.

B.  The actual yield is 20 g.  This information is given in the problem.

C.  Since you know how much product you will get, convert moles of Na₃AlF₆ to grams to find actual yield.  Use the value found for Na₃AlF₆ that you got from the limiting reagent.  The molar mass is 209.94 g/mol.

(0.0875 mol Na₃AlF₆) × (209.94 g/mol Na₃AlF₆) = 18.4 g Na₃AlF₆

D.  To find the percent yield, divide the actual yield by the theoretical yield  and multiply by 100%.

(20 g)/(18.4 g) × 100% = 108%

If your percent yield is greater than 100% when performing a reaction, there has been a mistake somewhere.  Either you recorded the numbers incorrectly or there has been some human error in your experiment.  This might be the right answer for this problem, but you might want to double check to make sure the numbers you gave me were right.

what's the

i think the answer is terrae but not 100 percent sure. ▓▬▬ §