Calcium oxide and carbon are reacted to produce carbon monoxide and calcium carbide CaC2. When one mole of calcium carbide is formed, 464.8kJ is absorbed.
What is the enthlapy of reaction when 1.00g of calcium carbide is formed?
See your post above.
To calculate the enthalpy of reaction when 1.00g of calcium carbide (CaC2) is formed, we need to use the given information about the enthalpy change (ΔH) when one mole of calcium carbide is formed.
First, we need to convert the given mass of calcium carbide to moles. We can do this by using the molar mass of calcium carbide.
The molar mass of calcium (Ca) is 40.08 g/mol, and the molar mass of carbon (C) is 12.01 g/mol. The molar mass of calcium carbide (CaC2) can be calculated as:
Molar mass of CaC2 = (1 × molar mass of Ca) + (2 × molar mass of C)
= (1 × 40.08 g/mol) + (2 × 12.01 g/mol)
= 64.10 g/mol
Next, we need to calculate the number of moles of calcium carbide using the given mass:
Number of moles of CaC2 = mass of CaC2 / molar mass of CaC2
= 1.00 g / 64.10 g/mol
Now that we have the number of moles of calcium carbide, we can calculate the enthalpy change when 1.00g of calcium carbide is formed. The enthalpy change (ΔH) is given as 464.8 kJ when one mole of calcium carbide is formed. Therefore, we can use the following equation:
Enthalpy change = ΔH / number of moles of CaC2
Let's calculate it:
Enthalpy change = 464.8 kJ / (1.00 g / 64.10 g/mol)
= 464.8 kJ / (0.0156 mol)
= 29,827 kJ/mol (rounded to the nearest whole number)
Therefore, the enthalpy change when 1.00g of calcium carbide is formed is approximately 29,827 kJ/mol.