Use the chemical reaction to answer the question.
CaCO3→CO2+CaO
A scientist heats 30 grams of calcium carbonate (CaCO3) in the chemical reaction shown. What is the total mass of the products?
Remember the Law of conservation of matter; ie., matter is neither created nor destroyed.
30 grams
To determine the total mass of the products in the chemical reaction, we need to understand the stoichiometry of the reaction.
According to the balanced equation:
1 mole of CaCO3 reacts to produce 1 mole of CO2 and 1 mole of CaO.
First, we need to calculate the number of moles of CaCO3:
Molar mass of CaCO3 = atomic mass of Ca + (atomic mass of C + (3 * atomic mass of O))
= 40.08 g/mol + (12.01 g/mol + (3 * 16.00 g/mol))
= 40.08 g/mol + 12.01 g/mol + 48.00 g/mol
= 100.09 g/mol
Number of moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3
= 30 g / 100.09 g/mol
= 0.299 moles of CaCO3
Since the stoichiometry of the reaction is 1:1, the number of moles of CO2 and CaO will also be 0.299 moles.
Now, let's calculate the mass of each product:
Mass of CO2 = number of moles of CO2 * molar mass of CO2
= 0.299 moles * 44.01 g/mol (molar mass of CO2)
= 13.17 g
Mass of CaO = number of moles of CaO * molar mass of CaO
= 0.299 moles * 56.08 g/mol (molar mass of CaO)
= 16.76 g
Therefore, the total mass of the products (CO2 + CaO) is 13.17 g + 16.76 g = 29.93 g.
To find the total mass of the products in the given chemical reaction, we can use stoichiometry. Stoichiometry is a way to quantitatively relate the mass of reactants to the mass of products in a chemical reaction.
In this reaction, the reactant is calcium carbonate (CaCO3) and the products are carbon dioxide (CO2) and calcium oxide (CaO). The balanced equation gives us the stoichiometric coefficients, which represent the mole ratios between the reactant and products.
First, we need to calculate the moles of calcium carbonate (CaCO3) using its molar mass. The molar mass of calcium carbonate can be found by summing the atomic masses of its constituent elements: Ca = 40.08 g/mol, C = 12.01 g/mol, and O = 16.00 g/mol.
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + 3(16.00 g/mol) = 100.09 g/mol
To convert grams of calcium carbonate to moles, we use the equation:
moles = mass / molar mass
moles of CaCO3 = 30 g / 100.09 g/mol = 0.2999 mol (approximately)
Now, using the stoichiometric coefficients from the balanced equation, we can determine the moles of products formed.
From the balanced equation:
1 mol CaCO3 produces 1 mol CO2
1 mol CaCO3 produces 1 mol CaO
Therefore, the moles of CO2 and CaO will also be 0.2999 mol.
Finally, we find the mass of the products by multiplying the moles by their respective molar masses:
mass of CO2 = moles of CO2 * molar mass of CO2 = 0.2999 mol * 44.01 g/mol = 13.20 g (approximately)
mass of CaO = moles of CaO * molar mass of CaO = 0.2999 mol * 56.08 g/mol = 16.81 g (approximately)
Hence, the total mass of the products in the chemical reaction is approximately 13.20 g (CO2) + 16.81 g (CaO) = 29.01 g.