What mass of limestone will produce 11.2g of lime
Set up your balanced equation.
How many moles in 11.2g of lime?
Use the equation to figure the moles used
Convert that to grams
Well, that depends on the molar mass of limestone and lime. But let's not get too serious here. Why did the limestone go to the party? Because it wanted to have a smashing time and turn into lime! Get it? Smashing? Okay, okay, I'll give you a more serious answer now. The molar mass of limestone (CaCO3) is approximately 100.0869 g/mol, and the molar mass of lime (CaO) is approximately 56.0774 g/mol. So if we assume 100% yield, you would need (11.2 g lime) * (100.0869 g limestone / 56.0774 g lime) = 19.99 g of limestone. But remember, actual yields can vary, so don't be too carbonate about it!
To find the mass of limestone that will produce 11.2g of lime, we need to consider the chemical equation for the reaction between limestone (calcium carbonate) and heat to produce lime (calcium oxide).
The balanced chemical equation for this reaction is:
CaCO3 -> CaO + CO2
From the equation, we can see that one mole of calcium carbonate (CaCO3) produces one mole of calcium oxide (CaO).
Now, let's calculate the molar mass of calcium oxide (CaO) and calcium carbonate (CaCO3):
- Molar mass of CaO = 40.08 g/mol (calcium: 40.08 g/mol, oxygen: 16.00 g/mol)
- Molar mass of CaCO3 = 100.09 g/mol (calcium: 40.08 g/mol, carbon: 12.01 g/mol, oxygen: 16.00 g/mol multiplied by 3)
To find the mass of limestone needed, we can set up a proportion using the molar mass:
mass of calcium carbonate (g) / molar mass of CaCO3 (g/mol) = mass of lime (g) / molar mass of CaO (g/mol)
Let's plug in the given mass of lime (11.2g) into the proportion:
mass of calcium carbonate (g) / 100.09 g/mol = 11.2g / 40.08 g/mol
Cross-multiply and solve for the mass of calcium carbonate:
mass of calcium carbonate (g) = (11.2g / 40.08 g/mol) * 100.09 g/mol
mass of calcium carbonate (g) ≈ 31.35g
Therefore, approximately 31.35 grams of limestone will produce 11.2g of lime.
To determine the mass of limestone required to produce 11.2g of lime, we need to know the chemical equation for the reaction involved. The reaction between limestone (calcium carbonate, CaCO3) and heat (thermal decomposition) produces lime (calcium oxide, CaO) and carbon dioxide (CO2).
The balanced chemical equation for this reaction is:
CaCO3 → CaO + CO2
From the equation, we can see that 1 mole of limestone (CaCO3) yields 1 mole of lime (CaO). To find the mass of limestone required, we can use the molar mass of CaCO3, which is approximately 100.09 g/mol.
Let's calculate the mass of limestone required:
1. Determine the number of moles of lime produced:
The molar mass of lime (CaO) is approximately 56.08 g/mol. Since 11.2g of lime is given, we can calculate the number of moles using the formula:
Moles = Mass / Molar mass
= 11.2g / 56.08 g/mol
≈ 0.2 moles
2. The molar ratio between limestone (CaCO3) and lime (CaO) is 1:1. This means that 1 mole of limestone produces 1 mole of lime.
Therefore, the number of moles of limestone required is also 0.2 moles.
3. Calculate the mass of limestone required:
Mass = Moles x Molar mass
= 0.2 moles x 100.09 g/mol
≈ 20.02 g
Therefore, approximately 20.02 grams of limestone will produce 11.2 grams of lime.