1a. Assuming the experiment was executed without error, 16. Use the starting amount of Mg,0.390g, and the result from determine the mass of MgO produced from 0.390g of Mg by the part (1a) to determine the mass percent of Mg in MgO. reaction: 2Mg(s)+O2(g)2MgO(s) using that 1.6583g of MgO is formed from 1.0000gMg. Enter the result in g to the nearest 0.001g. Do not enter the Enter the result in percent to the nearest 0.01%. Do not enter unit. the % symbol. In the experiment, water is added to convert the small amount of magnesium nitride (Mg3N2) that might have been produced during magnesium (Mg) combustion into magnesium oxide (MgO). This conversion takes place in two chemical steps. Mg3N2(s)+6H2O(g)3Mg(OH)2(s)+2NH3(g)Mg(OH)2(s)heatMgO(s)+H2O(g) Suppose that the second step does not take place and answer the following questions assuming that any Mg that reacted to form Mg3N2 remains in the products as magnesium hydroxide (Mg(OH)2). Consider the same 0.390g of magnesium (Mg) is heated in a crucible under combustion and subsequent conditions where 89.51% of the magnesium is converted into magnesium oxide (MgO) and 10.49% of the magnesium is ultimatoly converted into magnesium hydroxide (Mg(OH)2). 2e. If the experimental results were analyzed as if the mass of all solid products (MgO(s) and Mg(OH)2(s) was pure MgO, what mass percent of Mg in MgO would be determined? Enter the result to the nearest 0.01%. Do not enter the % symbol. 3. Select the words which correctly summarizes the result of the procedural error. HINT. Compare the answers to part 1 and part 20. The determine mass percent of Mg in MgO is because the mass percent of Mg is in M9O