If 28.8 grams of calcium carbonate react to form 16.1 grams of calcium oxide, how many grams of carbon dioxide must simultaneously be formed?
molar Mass CaCO3 = 40+12+48 = 100 g/mol
CaO = 40 + 16 = 56 g/mol
CO2 = 12+32 = 44 g/mol
so
28.8 (44/100) = 12.7 grams
To answer this question, we need to use the balanced chemical equation for the reaction between calcium carbonate (CaCO3) and calcium oxide (CaO):
CaCO3 → CaO + CO2
Looking at the equation, we can see that for every 1 mole of calcium carbonate, 1 mole of calcium oxide and 1 mole of carbon dioxide are produced.
First, we need to convert the given masses of calcium carbonate and calcium oxide into moles.
The molar mass of calcium carbonate (CaCO3) is:
40.08 g/mol (Ca) + 12.01 g/mol (C) + (3 * 16.00 g/mol) (O) = 100.09 g/mol
Using the given mass of calcium carbonate (28.8 g), we can calculate the number of moles:
28.8 g / 100.09 g/mol = 0.288 moles
The molar mass of calcium oxide (CaO) is:
40.08 g/mol (Ca) + 16.00 g/mol (O) = 56.08 g/mol
Using the given mass of calcium oxide (16.1 g), we can calculate the number of moles:
16.1 g / 56.08 g/mol = 0.287 moles
Now that we know the number of moles of calcium carbonate and calcium oxide, we can determine the number of moles of carbon dioxide produced using the mole ratios from the balanced equation.
Since the balanced equation tells us that the ratio of calcium carbonate to carbon dioxide is 1:1, then the moles of carbon dioxide produced is also 0.288 moles.
Finally, we can convert the moles of carbon dioxide to grams using the molar mass of carbon dioxide (CO2):
Molar mass of CO2 = 12.01 g/mol (C) + (2 * 16.00 g/mol) (O) = 44.01 g/mol
Multiplying the moles of carbon dioxide by the molar mass, we get:
0.288 moles * 44.01 g/mol = 12.67 grams
Therefore, 12.67 grams of carbon dioxide must be simultaneously formed.
To determine the number of grams of carbon dioxide formed when 28.8 grams of calcium carbonate react, we need to use stoichiometry and the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between calcium carbonate (CaCO3) and calcium oxide (CaO) is:
CaCO3 -> CaO + CO2
We can see from the balanced equation that for every 1 mole of calcium carbonate reacted, 1 mole of calcium oxide and 1 mole of carbon dioxide are formed.
To solve this problem, we need to convert the given mass of calcium carbonate (28.8 grams) into moles. To do this, we divide the mass by the molar mass of calcium carbonate, which can be calculated by adding up the atomic masses of calcium (Ca), carbon (C), and oxygen (O).
The atomic masses are:
Ca: 40.1 g/mol
C: 12.0 g/mol
O: 16.0 g/mol
So, the molar mass of CaCO3 is:
40.1 g/mol + 12.0 g/mol + (16.0 g/mol x 3) = 100.1 g/mol
Now we can calculate the moles of calcium carbonate:
Moles of CaCO3 = 28.8 g / 100.1 g/mol
Next, we use the mole ratio from the balanced equation to find the moles of carbon dioxide formed. From the balanced equation, we know that 1 mole of calcium carbonate produces 1 mole of carbon dioxide.
Moles of CO2 = Moles of CaCO3
Finally, we convert the moles of carbon dioxide back into grams by multiplying by the molar mass of carbon dioxide. The molar mass of CO2 is:
C: 12.0 g/mol
O: 16.0 g/mol x 2 = 32.0 g/mol
Moles to grams: Moles of CO2 x molar mass of CO2
So, to calculate the grams of carbon dioxide produced, you need to follow these steps:
1. Calculate the moles of calcium carbonate by dividing the mass (28.8 g) by the molar mass of calcium carbonate (100.1 g/mol).
2. Use the mole ratio from the balanced equation to find the moles of carbon dioxide.
3. Convert the moles of carbon dioxide back into grams by multiplying by the molar mass of CO2 (44.0 g/mol).
By following these steps using the given values, you should be able to calculate the grams of carbon dioxide produced when 28.8 grams of calcium carbonate react.