One reaction of iron with hydrochloric acid is represented by the following

thermochemical equation.
Fe(s) + 2HCl(aq) ? FeCl2(aq) + H2(g); (Delta)H°= –87.9 kJ
How much heat is liberated at constant pressure if 0.215 g of iron reacts with 40.4 mL of
0.579 M HCl?
How much heat is liberated at constant pressure when 28 g of calcium oxide reacts with 86.5 L of carbon dioxide gas measured at 1 atm pressure and 25C( R=0.0821 L.atm\(K.mol) CaO(s)+CO2(g)---CaCO3(s);delta H=-178.3 kJ
These are final two HW problems I can not get the correct answer on either. Been waking on them for two days. Help show all steps so I can find my mistakes.

The better idea is for you to type and let us check. It's much faster that way. However, I'll split it with you and do the first.

mols Fe = grams/molar mass. (You need to check that Fe is the limiting reagent.(It is.)
Basically you read it this way. If 87.9 kJ heat is liberated for 1 mol (55.85) g Fe, how much is liberated for 0.215.
That's 87.9 kJ x (0.215/55.85) = ?kJ.

To find the amount of heat liberated in both reactions, we can use the equation:

q = (mole ratio)(∆H)

For the first reaction:
1. Calculate the moles of iron (Fe):
moles of Fe = mass of Fe / molar mass of Fe
moles of Fe = 0.215 g / 55.845 g/mol = 0.00385 mol

2. Calculate the moles of HCl:
moles of HCl = volume of HCl (L) x concentration of HCl (mol/L)
moles of HCl = 0.0404 L x 0.579 mol/L = 0.0234 mol

3. Determine the mole ratio between Fe and HCl from the balanced equation:
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)
The mole ratio is 1:2 (1 mole of Fe reacts with 2 moles of HCl).

4. Calculate the heat liberated (q):
q = (mole ratio)(∆H)
q = (0.00385 mol Fe)(-87.9 kJ/1 mol) = -0.338 kJ (rounded to 3 decimal places)

Therefore, the amount of heat liberated at constant pressure is approximately -0.338 kJ.

For the second reaction:
1. Calculate the moles of calcium oxide (CaO):
moles of CaO = mass of CaO / molar mass of CaO
moles of CaO = 28 g / 56.077 g/mol = 0.4996 mol

2. Convert the volume of carbon dioxide gas to moles:
moles of CO2 = volume of CO2 (L) / 22.414 L/mol
moles of CO2 = 86.5 L / 22.414 L/mol = 3.8648 mol

3. Determine the mole ratio between CaO and CO2 from the balanced equation:
CaO(s) + CO2(g) → CaCO3(s)
The mole ratio is 1:1 (1 mole of CaO reacts with 1 mole of CO2).

4. Calculate the heat liberated (q):
q = (mole ratio)(∆H)
q = (0.4996 mol CaO)(-178.3 kJ/1 mol) = -89.2 kJ (rounded to 3 decimal places)

Therefore, the amount of heat liberated at constant pressure is approximately -89.2 kJ.

To solve these problems, we can use the concept of stoichiometry and the given values of reactants to calculate the heat liberated at constant pressure. Here's how you can approach each problem:

Problem 1:
1. Convert the mass of iron (Fe) to moles by using the molar mass of iron (55.85 g/mol).
Moles of Fe = 0.215 g / 55.85 g/mol = 0.00385 mol

2. Convert the volume of hydrochloric acid (HCl) to moles by using the molarity of HCl (0.579 M).
Moles of HCl = 0.0404 L * 0.579 mol/L = 0.0234 mol

3. Use the balanced equation to determine the mole ratio between Fe and HCl.
From the equation: 1 mol Fe reacts with 2 mol HCl

4. Determine the limiting reactant.
Since there are fewer moles of Fe (0.00385 mol) compared to HCl (0.0234 mol), Fe is the limiting reactant.

5. Calculate the heat liberated using the molar enthalpy change (ΔH) value.
Heat liberated = moles of limiting reactant * ΔH
Heat liberated = 0.00385 mol * (-87.9 kJ/mol) = -0.338 kJ (negative sign indicates heat release)

Therefore, 0.338 kJ of heat is liberated at constant pressure.

Problem 2:
1. Convert the mass of calcium oxide (CaO) to moles by using the molar mass of CaO (56.08 g/mol).
Moles of CaO = 28 g / 56.08 g/mol = 0.499 mol

2. Convert the volume of carbon dioxide (CO2) gas to moles by using the ideal gas law equation: PV = nRT.
Moles of CO2 = (86.5 L * 1 atm) / (0.0821 L.atm/(K.mol) * 298 K) = 3.42 mol

3. Use the balanced equation to determine the mole ratio between CaO and CO2.
From the equation: 1 mol CaO reacts with 1 mol CO2

4. Determine the limiting reactant.
Since the stoichiometric ratio is 1:1 between CaO and CO2, their moles are the same. Therefore, neither is a limiting reactant in this case.

5. Calculate the heat liberated using the molar enthalpy change (ΔH) value.
Heat liberated = moles of reactant * ΔH
Heat liberated = 0.499 mol * (-178.3 kJ/mol) = -88.9 kJ (negative sign indicates heat release)

Therefore, 88.9 kJ of heat is liberated at constant pressure.

Make sure to double-check the calculations and account for any unit conversions. Let me know if you need further assistance or clarification!