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:

Trial 1: 0.423 gTrial 2: 0.585 g

Question 2:

Trial 1: 98.9%Trial 2: 98.4%

Question 3:

98.7%

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:

number of moles of Mg: 0.255g/24.305g/mol = 0.01049 mol

Trial 2:

number of moles of Mg: 0.353g/24.305g/mol = 0.01452 mol

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:

Trial 1: 0.01049  mol MgOTrial 2: 0.01452 mol MgO

c) Convert the number of moles to mass:

i) Formula:

mass = number of moles × molar mass =

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:

Percent yield = (actual yield / theoretical yield) × 100

b) Trial 1

% yield = (0.418g / 0.423g) × 100 = 98.9%

c) Trial 2:

% yield = (0.576g / 0.585g) × 100 = 98.4%

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:

Average % yield = [98.9% + 98.4%]/2 = 98.6%

Hno3(aq) → h+(aq) + no3−(aq). picture ii best represents the strong acid hno3. strong acids are strong electrolytes

The answer is 65.3312