Question

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3 pages.

Chem 143 - Lab Pre-lab Exercise Show the details of each calculation to get full credit 1. Magnesium oxide, a white powdery substance can often be made by heating magnesium metal in presence of oxygen. a. Write the balanced chemical equation for this reaction (including the state symbols such as (s) () or (aq)] b. When 10.1 g magnesium is allowed to react with 10.5g of oxygen gas, 11.9 g of magnesium oxide are collected. Determine the limiting reactant, theoretical yield and percent yield for the reaction

Answer 1

a) The balanced chemical reaction of Mg with O2 is given by:-

2 Mg(s) + O2(g) ---> 2 MgO(s)

b) To find out the limiting reactant, theoretical and percent yield first we need to find moles of each reactant. Then by using mole to mole ratio we need to find moles of product for each reactant and we can do this in one step using dimensional analysis.

Since mole to mole ratio for Mg to MgO is 2:2.

Therefore using dimensional analysis,

10.1 g Mg * (1 mol Mg/24.31 g Mg) * (2 mol MgO/2 mol Mg) = 20.2 mol MgO/24.31 = 0.830933 mol MgO

Mole to mole ratio for O2 to MgO is 1:2.

10.5 g O2 * (1 mol O2/32 g O2) * (2 mol MgO/1 mol O2) = 21 mol MgO/32 = 0.65625 mol MgO

Since the oxygen prodcues less number of mole for product. Therefore, limiting reactant for above reaction is O2.

To find the theoretical yiled, use moles of limiting reactant which is 0.65625.

Theoretical yield = 0.65625 mol MgO * (40.3044 g MgO/1 mol MgO) = 26.2725 g MgO = 26.3 g MgO

Percent yield = (Actual Yield/Theoretical Yield)*100

= (11.9 g/26.3 g) * 100 = 0.45247 *100 = 45.247% = 45.25% = 45.2% yield.