1. Write the balanced chemical equation for the reaction you are performing. 2. Subtract the mass of the crucible and lid (row 1 in the chart) from the total mass of Mg, crucible, and lid (row 2 in the chart) to find the mass of magnesium for each trial.
• Trial 1: 0.252
• Trial 2: 0.406

3. Subtract the mass of the crucible and lid (row 1 in the chart) from the total mass of MgO, crucible, and lid (row 3 in the chart) to find the mass of magnesium oxide for each trial. This is the actual yield of magnesium oxide for each trial.
• Trial 1: 0.413
• Trial 2: 0.459

4. Magnesium is the limiting reactant in this experiment. Calculate the theoretical yield of MgO for each trial.
• Trial 1:
• Trial 2:

5. Determine the percent yield of MgO for your experiment for each trial.
• Trial 1:
• Trial 2:
6. Determine the average percent yield of MgO for the two trials.

3

Explanation:

because

Explanation:

youu do this then you do that and then you do this that and this

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml