Step-by-step explanation:
amount of baking soda = 2.01 g
---> 2.01 g NaHCO3 × (1 mol NaHCO3/84.007 g NaHCO3)
= 0.024 mol NaHCO3
mass of the product = 1.28 g
Possible reactions:
NaHCO3(s) ---> NaOH(s) + CO2(g)
0.024 mol NaHCO3 × (1 mol NaOH/1 mol NaHCO3)
= 0.024 mol NaOH × (39.997 g NaOH/1.mol NaOH)
= 0.96 g NaOH
2NaHCO3(s) ---> Na2O(s) + 2CO2(g) + H2O(g)
0.024 mol NaHCO3 × (1 mol Na2O/2 mol NaHCO3)
= 0.012 mol Na2O × (61.9789 g Na2O)
= 0.73 g Na2O
2NaHCO3(s) ---> Na2CO3(s) + CO2(g) + H2O(g)
0.024 mol NaHCO3 × (1 mol Na2CO3/2 mol NaHCO3)
= 0.012 mol Na2CO3 × (105.9888 g Na2CO3/1 mol Na2CO3)
= 1.27 g Na2CO3
Conclusion:
Based on the mass of the product in the crucible and comparing it with the predicted product mass in different reactions, I conclude that Na2CO3 is the most likely product in the crucible.