PLEASE HELP!!

Answer the following questions using the data below:

Data :Trial 1 :Trial 2

Mass of empty crucible with lid: 26.679 grams 26.698 grams

Mass of Mg metal, crucible, and lid: 26.934 grams 27.051 grams

Mass of MgO, crucible, and lid: 27.097 grams 27.274 grams

Balanced Chemical Equation for reaction: 2 MG(s) + O2(g) = 2 MGO(s)

Mass of magnesium for each trial:

Trial 1: 0.255g

Trial 2: 0.353g

Actual yield of magnesium oxide for each trial:

Trial 1: 0.418g

Trial 2: 0.576g

Question 1: Calculate the theoretical yield of MgO for each trial:

Question 2: Determine the percent yield of MgO for your experiment for each trial:

Question 3: Determine the average percent yield of MgO for the two trials:

Question 1:

Question 2:

Question 3:

Explanation:

1. Organize all the data in one table:

Data:

                                                           (masses in grams)

                                                             Trial 1      Trial 2

Mass of empty crucible with lid:          26.679       26.698  

Mass of Mg metal, crucible, and lid:    26.934       27.051

Mass of MgO, crucible, and lid:            27.097        27.274

Mass of Mg metal                                 0.255            0.353

Actual yield of MgO                             0.418g           0.576

2. Question 1: Calculate the theoretical yield of MgO for each trial

a) Balanced chemical equation

b) Mole ratio

c) Convert the mass of Mg into number of moles

i) Formula: number of moles = mass in grams / molar mass

ii) Molar mass of Mg (same atomic mass): 24.305 g/mol

iii) Moles of Mg available:

Trtial 1:

Trial 2:

d) Use the mol ratio to calculate the moles of MgO

As per the stoichiometric ratio, one mol of Mg yield one mol of MgO. Thus:

c) Convert the number of moles to mass:

i) Formula:

Trial 1: 0.01049 mol × 40.3044 g/mol = 0.4229g = 0.423g

Trial 2: 0.01452 mol × 40.3044 g/mol = 0.5854g = 0.585 g

Those are the theoretical yields fore ach trial: 0.423g and 0.585 g

3. Question 2. Determine the percent yield of MgO for each trial

a) Formula:

b) Trial 1

c) Trial 2:

4. Question 3. Determine the average percent yield of MgO for the two trials

You can add the two percent yields and divide by 2:

increased stability in the completion of fluorine's outer shell.