To determine the total mass of the products in the given chemical reaction, we first need to understand the stoichiometry of the reaction.
The balanced equation for the reaction is:
CaCO3 → CO2 + CaO
According to the equation, one mole of calcium carbonate (CaCO3) will produce one mole of carbon dioxide (CO2) and one mole of calcium oxide (CaO).
The molar mass of calcium carbonate (CaCO3) is:
Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Adding these up, we get:
40.08 g/mol + 12.01 g/mol + (3 × 16.00 g/mol) = 100.09 g/mol
Therefore, 1 mole of calcium carbonate (CaCO3) has a mass of 100.09 grams.
Now, we can calculate the molar masses of carbon dioxide (CO2) and calcium oxide (CaO):
CO2: 12.01 g/mol + (2 × 16.00 g/mol) = 44.01 g/mol
CaO: 40.08 g/mol + 16.00 g/mol = 56.08 g/mol
Since we're given 30 grams of calcium carbonate (CaCO3), we can calculate the number of moles of CaCO3:
30 grams / 100.09 g/mol = 0.2999 moles (approximately 0.3 moles)
Using the stoichiometry of the reaction, we know that for every mole of CaCO3, we obtain one mole of CO2 and one mole of CaO.
Therefore, the total moles of products would also be 0.3 moles of CO2 and 0.3 moles of CaO.
Now, we can calculate the masses of the products:
Mass of CO2 = 0.3 moles × 44.01 g/mol = 13.20 grams
Mass of CaO = 0.3 moles × 56.08 g/mol = 16.82 grams
Adding up the masses of the products:
Mass of products = Mass of CO2 + Mass of CaO = 13.20 grams + 16.82 grams = 30.02 grams
Therefore, the total mass of the products is approximately 30.02 grams.