TAMUK Chemistry Department 1311 General Chemistry Your name Mid-term Assessment
ID: 1087817 • Letter: T
Question
TAMUK Chemistry Department 1311 General Chemistry Your name Mid-term Assessment Exam Your K# Total points: 250 3 In a preparation of CaO, 2.75 mol of Ca reacts with 2.85 mol of O. (a) Define and find the limiting reactant (6 pts) b) Calculate the product yield in grams. (14 pts) Define the limiting reactant: Point 1 Point 2 Point 3 Balanced reaction Ca(s) + O2(g) CaO(s) Moles of CaO calucated from Ca Moles of Cao calucated from O2 Comparison of the moles of Cao produced from both reactants, Ca and O2. What is the limiting reactant in this reaction? Molar mass of caCa o o 78 o O 15.1243/ mel Masof Cao0723Explanation / Answer
Balanced equation is,
2 Ca + O2 -----------> 2 CaO
The reactant which consumes completely in the reaction and decides the yield of product is called limiting reagent.
(a)
from the balanced equation,
2 mol of Ca forms 2 mol of CaO
Then,
2.75 mol of Ca froms 2.75 mol of CaO
(b)
From the balanced equation,
1 mol of O2 forms 2 mol of CaO
then,
2.85 mol of O2 forms = 5.70 mol of CaO
(c)
With respect to calcium we get less number of moles of CaO
(d)
Hence Ca is limiting reagent.
(e)
Molar mass of CaO = 40 + 16 = 56 g/mol
(f)
Mass of CaO = moles * molar mass = 2.75 * 56 = 154 g.
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