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?
According to the chemical reaction, 1 mole of CaCO3 produces 1 mole of CO2 and 1 mole of CaO.
The molar mass of CaCO3 is 100.0869 g/mol.
Therefore, the number of moles of CaCO3 in 30 grams is:
30 g / 100.0869 g/mol = 0.2997 mol of CaCO3
Since the mole ratio is 1:1:1, the number of moles of CO2 and CaO produced is also 0.2997 mol each.
The molar mass of CO2 is 44.01 g/mol and the molar mass of CaO is 56.0774 g/mol.
Therefore, the total mass of the products is:
(0.2997 mol of CO2) x (44.01 g/mol) + (0.2997 mol of CaO) x (56.0774 g/mol)
= 13.199 g + 16.814 g
= 29.013 g
The total mass of the products is approximately 29.013 grams.
That’s incorrect
I apologize for the incorrect response. Let's go through the calculation again:
The molar mass of CaCO3 is 100.0869 g/mol. Therefore, 30 grams of CaCO3 is equal to:
30 g / 100.0869 g/mol = 0.2997 mol of CaCO3
According to the chemical reaction, 1 mole of CaCO3 produces 1 mole of CO2 and 1 mole of CaO. So, the number of moles of CO2 and CaO produced is also 0.2997 mol each.
The molar mass of CO2 is 44.01 g/mol and the molar mass of CaO is 56.0774 g/mol.
Therefore, the total mass of the products is:
(0.2997 mol of CO2) x (44.01 g/mol) + (0.2997 mol of CaO) x (56.0774 g/mol)
= 13.18 g + 16.81 g
= 29.99 g
So, the correct total mass of the products is approximately 29.99 grams.
According to the balanced chemical equation:
1 mol of CaCO3 produces 1 mol of CO2 and 1 mol of CaO.
The molar mass of CaCO3 is:
(40.08 g/mol for Ca) + (12.01 g/mol for C) + (3 * 16.00 g/mol for O) = 100.09 g/mol
Given that there are 30 grams of CaCO3, we can determine the number of moles by dividing the mass by the molar mass:
30 g / (100.09 g/mol) = 0.2999 mol (rounded to 4 decimal places)
Since the molar ratio between CaCO3 and the products is 1:1, we would expect the same number of moles for CO2 and CaO.
Therefore, the total mass of the products is:
0.2999 mol x (44.01 g/mol for CO2 + 56.08 g/mol for CaO) = 29.99 grams.
The total mass of the products is approximately 29.99 grams.
To determine the total mass of the products in the chemical reaction, we need to consider the law of conservation of mass, which states that the mass of the reactants must be equal to the mass of the products.
In this particular reaction, 1 mole of calcium carbonate (CaCO3) reacts to produce 1 mole of carbon dioxide (CO2) and 1 mole of calcium oxide (CaO).
To calculate the total mass of the products, we first need to convert the given mass of calcium carbonate (30 grams) into moles. To do this, we divide the given mass by the molar mass of calcium carbonate.
- The molar mass of calcium carbonate (CaCO3):
- The atomic mass of calcium (Ca) is 40.08 g/mol.
- The atomic mass of carbon (C) is 12.01 g/mol.
- The atomic mass of oxygen (O) is 16.00 g/mol.
Adding these atomic masses together, we get:
(1 x Ca) + (1 x C) + (3 x O) = 40.08 + 12.01 + (3 x 16.00) = 40.08 + 12.01 + 48.00 = 100.09 g/mol
Thus, the molar mass of calcium carbonate is 100.09 g/mol.
Next, we calculate the number of moles of calcium carbonate by dividing the given mass of 30 grams by the molar mass of calcium carbonate:
30 g / 100.09 g/mol = 0.2999 moles (approximately 0.30 moles)
Since the reaction produces an equal number of moles of carbon dioxide and calcium oxide, we have 0.30 moles of each.
Finally, we multiply the number of moles by the molar mass of each product to determine their masses:
- Mass of carbon dioxide:
0.30 moles x molar mass of CO2 (44.01 g/mol) = 13.20 grams (approximately)
- Mass of calcium oxide:
0.30 moles x molar mass of CaO (56.08 g/mol) = 16.82 grams (approximately)
Therefore, the total mass of the products in the given reaction is approximately 13.20 grams (carbon dioxide) + 16.82 grams (calcium oxide) = 30.02 grams.