PLEASE HELP!! WILL MARK BRAINLIEST

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:

The answer to your question is below

Explanation:

                                         Trial 1                   Trial 2

mass of Mg                      0.255 g                0.353 g

mass of MgO                   0.418 g                 0.576 g

Chemical reaction

                                2Mg(s)    +  O₂(g)     ⇒   2MgO(s)

Question 1.

Atomic mass of Mg = 24.31 x 2 = 48.62 g

Molecular mass of MgO = 2(24.31 + 16) = 80.62 g

Trial 1

                          48.62 g of Mg 80.62 g of MgO

                          0.255 g               x

                          x = (0.255 x 80.62)/48.62

                          x = 0.422 g of MgO

Trial 2                48.62 g of Mg 80.62 g of MgO

                          0.353 g               x

                          x = (0.353 x 80.62)/48.62

                          x = 0.585 g of MgO

Question 2

Trial 1          

Percent yield = 0.418/0.422 x 100 = 99%

Trial 2

Percent yield = 0.576/0.585 x 100 = 98.5%

Question 3

Average = (99 + 98.5)/2

              = 98.75%

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'