#29. Magnesium oxide can be made by heating magnesium metal in the presence of t
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Question
#29. Magnesium oxide can be made by heating magnesium metal in the presence of the oxygen. The balanced equation for the reaction is 2Mg(s)+O2(g)?2MgO(s) When 10.2 g Mg is allowed to react with 10.4 g O2, 11.9 g MgO is collected.Part A) Determine the limiting reactant for the reaction: Mg (s).......... or.........(O2) (g)
Part B) Determine the theoretical yield for the reaction. Express your answer in grams.
Part C) Determine the percent yield for the reaction.Express your answer as a percent.
#29. Magnesium oxide can be made by heating magnesium metal in the presence of the oxygen. The balanced equation for the reaction is 2Mg(s)+O2(g)?2MgO(s) When 10.2 g Mg is allowed to react with 10.4 g O2, 11.9 g MgO is collected.
Part A) Determine the limiting reactant for the reaction: Mg (s).......... or.........(O2) (g)
Part B) Determine the theoretical yield for the reaction. Express your answer in grams.
Part C) Determine the percent yield for the reaction.Express your answer as a percent.
#29. Magnesium oxide can be made by heating magnesium metal in the presence of the oxygen. The balanced equation for the reaction is 2Mg(s)+O2(g)?2MgO(s) When 10.2 g Mg is allowed to react with 10.4 g O2, 11.9 g MgO is collected.
Part A) Determine the limiting reactant for the reaction: Mg (s).......... or.........(O2) (g)
Part B) Determine the theoretical yield for the reaction. Express your answer in grams.
Part C) Determine the percent yield for the reaction.Express your answer as a percent.
Explanation / Answer
Determine moles of reactants to see which (if any) is a limiting reagent. They stated it reacted with a certian amount of O2, they didn't say burned in air so you can't assume oxygen is present in excess!
10.2g/24.3g/mole = 0.420mole Mg
10.4g/32.0g/mole = 0.325moleO2
From the balanced equation, you can see that 2 Mg's react with each O2 so 0.325 mole of O2 is more than enough to convert all the Mg to MgO; Mg is the limiting reagent, so that is the number we need to use for the calculations.
Since the ratio of Mg reacted to MgO formed is 1 to 1 we would get 0.420 mole of MgO from 0.420 mole of Mg.
0.420mole x [24.3 + 16.0]g/mole = 16.9g
So the theoretical yield (most MgO you could recover if reaction was 100% complete and no losses in recovering etc) is 16.9g
The percent yield is based on the ratio of actual recovered yield to theoretical yield:
11.9g/16.9g x 100% = 70.41%
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