CaO(s) + 3C(s) > CaC2(s) + CO(g)
deltaHrxn = +464.8 kJ/mol-rxn
Is the reaction endothermic or exothermic? What is the enthalpy change if 10.0 g of CaO is allowed to react with an excess of carbon?
If delta Hrxn is +, the reaction is endothermic. Look at the problem. 464.8 kJ of heat is needed to react with (how much) CaO. That is 56 g. So the heat needed to convert 10.0 g must be
464.8 kJ x 10.0/56 = ??
exothermic and 82.9kj
Well, it looks like our good friend CaO gets together with C and decides to have a little party. The equation suggests that this reaction is endothermic because the deltaHrxn value is positive (+464.8 kJ/mol-rxn). So it's like this reaction needs a little extra heat to get things going.
Now, if you want to know the enthalpy change when 10.0 g of CaO is allowed to react with an excess of carbon, we'll have to put our math hats on. To do that, we need the molar mass of CaO. Since calcium has an atomic mass of 40.08 g/mol and oxygen has an atomic mass of 16.00 g/mol, the molar mass of CaO is 40.08 + 16.00 = 56.08 g/mol.
Now, we can figure out the number of moles of CaO in 10.0 g of it by dividing the mass by the molar mass. So, 10.0 g / 56.08 g/mol = 0.178 mol.
Since the reaction gives off +464.8 kJ of heat per mol-rxn, we can multiply this by the number of moles to find the heat change for 10.0 g of CaO. So, 464.8 kJ/mol-rxn * 0.178 mol = 82.71 kJ.
Therefore, if 10.0 g of CaO reacts with an excess of carbon, the enthalpy change will be approximately 82.71 kJ. Hope that sheds some light on this reaction for you!
The reaction CaO(s) + 3C(s) → CaC2(s) + CO(g) has a positive value for deltaHrxn (+464.8 kJ/mol-rxn). This indicates that the reaction is endothermic.
To determine the enthalpy change if 10.0 g of CaO reacts with an excess of carbon, we need to convert the mass of CaO to moles and then use the molar ratio from the balanced equation to find the corresponding enthalpy change.
First, calculate the number of moles of CaO:
Molar mass of CaO = 40.08 g/mol
Moles of CaO = (mass of CaO) / (molar mass of CaO)
Moles of CaO = 10.0 g / 40.08 g/mol
Next, use the molar ratio from the balanced equation:
From the balanced equation: CaO(s) + 3C(s) → CaC2(s) + CO(g)
The stoichiometric ratio is 1:3 for CaO:C, which means for every 1 mole of CaO, we have 3 moles of C.
Finally, use the molar ratio to find the enthalpy change:
Enthalpy change = (moles of CaO) × (deltaHrxn)
Enthalpy change = (moles of CaO) × (+464.8 kJ/mol-rxn)
Simply substitute the calculated moles of CaO into the equation to find the enthalpy change.
To determine whether the reaction is endothermic or exothermic, we can use the value of the enthalpy change (ΔHrxn). If ΔHrxn is positive, the reaction is endothermic (absorbs heat from the surroundings). If ΔHrxn is negative, the reaction is exothermic (releases heat into the surroundings).
Given that ΔHrxn = +464.8 kJ/mol-rxn, we can conclude that the reaction is endothermic.
To find the enthalpy change when 10.0 g of CaO is allowed to react with an excess of carbon, we need to convert grams into moles.
First, we need to determine the number of moles of CaO:
Molar mass of CaO = 40.08 g/mol (atomic masses of Ca and O)
Number of moles of CaO = Mass of CaO / Molar mass of CaO
Number of moles of CaO = 10.0 g / 40.08 g/mol ≈ 0.2494 mol
Since there is an excess of carbon present, we can assume that all the CaO reacts. Therefore, the ΔHrxn can be calculated based on the number of moles of CaO.
Enthalpy change (ΔH) = ΔHrxn * Number of moles of CaO
Enthalpy change (ΔH) = +464.8 kJ/mol-rxn * 0.2494 mol
Enthalpy change (ΔH) ≈ +115.9 kJ
Therefore, the enthalpy change when 10.0 g of CaO reacts with an excess of carbon is approximately +115.9 kJ.