Question

Please answer questions 1-7 and fill out the equation at the top. Thanks!

LAB-Stoichiometry/Empirical formula Name Purpose: To predict the mass of magnesium oxide (MgO) formed by reacting To determine actual yield, theoretical yield and percent yield of pro Balance the equation: Moterials: Mg ribbon, Bunsen burner, balance, crucible tongs, crucible w/id etc- Procedure: place a check after you perform each step 1Go to the scale, check for zero. Record the mass of the crucible +lid and d it below 2. Curl the Mg into a coil, so that it fits in the bottom of the crucible. 3, Mass the magnesium ribbon, now place it back in the crucible uced from the sium. Get it Use the mass of the Mg ribbon to predict the mass of MgO that can above reaction. Assume that there is plenty of O, to completehy ren te data table checked by the teacher. Place predicted mass or theoretical yiela the K5. Place Mg in the crucible and put the lid on, leaving the lid slightl open. TURN ON THE BURNER The The ribbon should now be completely changed to a grey-w Weigh the crucible with the lid and magnesium oxide product to f !!! g will burn inside the crucible. DO NOT STARE AT THE BURNING MAGNESIUM rns out completely, cook it for an additional 5 minutes to assure the reaction is complete hite ash. 8. Turn off the heat. Let the crucible cool for 5 min 10. Subt 11. Return the clean crucible. Clean up materials and area before leaving. ind mass after heating- act the crucible with the lid weight from the ending mass to find mass of MgO (actua l yield) Data Table: Mass of crucible with (reactant) and crucièle with lid Mass of Mg Mass of Mgo product that must have Mass of axygen gas Mass of Mgo produced in Predicted/calculated mass experiment - actual Mgo product -theoretical lid after heating yield (accepted value) reacted yield (experimenta, value) 20.O19 449 20,84 2 69 Conclusions: 1. What was the mass of Mgo produced in the experiment? Did it match your predicted mass? 2. 3. Which yield can only be determined from an experiment? actual yield/theoretical yield/percent yield Use the experimental grams Mg and grams oxygen to calculate the empirical formula? Mg 4. Calculate the percent yield in your experiment. Is it possible to get a more than a percent yield theoretical yield actual yield-x100 100% yield? Explain. 5. Where did the O, that reacted with the Mg come from? Was there enough oxygen to react with the Mg? 6. Which reactant is the limiting reactant and which one is the excess reactant? 7. Calculate percent error. What are some possible errors in the experiment? percent error = accepted-experimental I × 100 accepted

Answer 1

R times n: 2 mg(s) + o_2(g) rightarrow 2 mgo(s) 2 mole of my require 1 mole of o_2 and give 2 mole of mgo 1 mole of my require 1/2 mole of o_2 and give 1 mole of mgo 0.018 mole of my require 0.016 times 1/2 mole of o_2 and give 0.018 mole of mgo required o_2 = 0.009 times 32 = 0.28 g mg o = product = 0.018 times 40 = 0.72 g 0.69 g mg o product in r times n do not match with predicted mass(0.72 g) Actual yield can only determined from experiment use experimental gram mg and grams oxygen to calculate enpirical formula mg: 0.45 g(0.018 mol) o: 0.28 g(0.009 mol) % yield = 0.69/0.72 times 100 = 95.8 % No it is not possible to get more than 100% yield because obtaining more than 100% yield means requiring more product than rectant it is only possible when your